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CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 12 - MCQExams.com
CBSE
Class 11 Engineering Maths
Sequences And Series
Quiz 12
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$$6$$
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$$7$$
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$$8$$
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$$9$$
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$$10$$
Find sum of series:
$$1.3.5+3.5.7+5.7.9.....$$ ?
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$$S_n=\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{8}+\frac{15}{8}$$
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$$S_n=\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{18}+\frac{15}{18}$$
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$$S_n=\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{20}+\frac{15}{8}$$
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$$S_n=\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{6}+\frac{1}{8}$$
$$\sin { \left( \dfrac { 1 }{ \sqrt { 2 } } \right) } +\sin { \left( \dfrac { \sqrt { 2 } -1 }{ \sqrt { 6 } } \right) } +\dots +\sin { \left( \dfrac { \sqrt { n } -\sqrt { n-1 } }{ \sqrt { n(n+1) } } \right) } +\dots \infty =$$
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$$\pi$$
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$$\dfrac {\pi}{2}$$
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$$\dfrac {\pi}{4}$$
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$$\dfrac {3\pi}{2}$$
The sum of the series $$1+\dfrac{1}{1!}.\dfrac{1}{4}+\dfrac{1\cdot 3}{2!}\left(\dfrac{1}{4}\right)^{2}+\dfrac{1\cdot 3 \cdot 5}{3!} \left(\dfrac{1}{4}\right)^{3}+........$$ to $$\infty$$ is ?
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$$\sqrt{2}$$
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$$2$$
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$$\dfrac{1}{\sqrt{2}}$$
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$$\sqrt{3}$$
The value of $$(21{C}_{1}-10{C}_{1})+(21{C}_{2}-10{C}_{2})+(3{C}_{1}-10{C}^{3})+(21{C}_{4}-10{C}^{4})+.....(21{C}_{10}-10{C}_{10})$$ is
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$${2}^{20}-{2}^{10}$$
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$${2}^{21}-{2}^{11}$$
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$${2}^{21}-{2}^{10}$$
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$${2}^{20}-{2}^{9}$$
If $${ a }_{ n }=\sum _{ r=0 }^{ n }{ \cfrac { 1 }{ { _{ }^{ n }{ C } }_{ r } } } $$, the value of $$\sum _{ r=0 }^{ n }{ \cfrac { n-2r }{ { _{ }^{ n }{ C } }_{ r } } } $$
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$$\cfrac{n}{2}{a}_{n}$$
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$$\cfrac{1}{2}{a}_{n}$$
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$$n{a}_{n}$$
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$$0$$
The value of $$\displaystyle\sum^{100}_{r=2}\dfrac{3^r(2-2r)}{(r+1)(r+2)}$$ is equal to?
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$$\dfrac{1}{2}-\dfrac{3^{100}}{100(101)}$$
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$$\dfrac{3}{2}-\dfrac{3^{101}}{101(102)}$$
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$$\dfrac{3}{2}-\dfrac{3^{100}}{100(101)}$$
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None of these
Explanation
$$ \displaystyle S= \sum_{r = 2}^{100} \dfrac{3^{r}(2-2r)}{r(r+1)(r+2)}$$
$$ \displaystyle S=\sum_{r = 2}^{100} \dfrac{3^{r}}{r(r+2)}-\dfrac{3^{r}+1}{(r+1)(r+2)}$$
$$ = \dfrac{3^{2}}{2(3)} - \dfrac{3^{3}}{3 \times 4}+ \dfrac{3^{3}}{3 \times 4} - \dfrac{3^{4}}{4\times 5}+\dfrac{3^{4}}{4\times 5}+...+\left ( \dfrac{3^{100}}{100 \times 101}-\dfrac{3^{101}}{101 \times 102} \right ) $$
$$ = \dfrac{3^{2}}{2(3)} - \dfrac{3^{101}}{101 \times 102 }$$
$$ S = \dfrac{3}{2} - \dfrac{3^{101}}{101 \times 102}$$
The sum of $$n$$ terms of the series $${1^3} + {3^3} + {5^3} + \ldots $$ is:
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$$3n^2(2n^2-1)$$
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$$n^2(2n^2-1)$$
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$$n^2(n^2-1)$$
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$$n^2(2n^2-5)$$
If $$f(x)=\displaystyle\Pi^3_{i=1}(x-a_i)+\displaystyle\sum^3_{i=1}a_i-3x$$ where $$a_i < a_{i+1}$$ for $$i=1, 2,$$ then $$f(x)=0$$ has:
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Only one distinct real root
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Exactly two distinct real root
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Exactly $$3$$ distinct real roots
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$$3$$ equal real roots
The sum of $$ \dfrac { 7 }{ 2\times 3 } \left( \dfrac { 1 }{ 3 } \right) +\dfrac { 9 }{ 3\times 4 } { \left( \dfrac { { 1 } }{ 3 } \right) }^{ 2 }+\dfrac { 11 }{ 4\times 5 } { \left( \dfrac { 1 }{ 3 } \right) }^{ 3 }+$$ upto 10 terms is equal to
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$$ \dfrac { 1 }{ 2 } -\dfrac { 1 }{ { 12\times }3^{ 10 } } $$
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$$ \dfrac { 1 }{ 3 } -\dfrac { 1 }{ { 12\times }3^{ 10 } } $$
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$$ \dfrac { 1 }{ 2 } -\dfrac { 1 }{ { 10\times }3^{ 10 } } $$
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$$None\ of\ these$$
The sum of the series
$$(2)^2+2(4)^{2}+3(6)^{2}+....$$ upto $$10$$ terms is
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$$12100$$
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$$11300$$
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$$11200$$
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$$12300$$
Explanation
$${\left(2\right)}^{2}+2{\left(4\right)}^{2}+3{\left(6\right)}^{2}+…..upto 10 term$$
$${T}_{m}=n{\left(2n\right)}^{2}\Rightarrow 4{n}^{3}$$
$$\sum_{n=0}^{10}{{T}_{n}}=4\sum_{n=0}^{10}{{n}^{3}}=\dfrac{4{\left(n\right)}^{2}{\left(n+1\right)}^{2}}{4}$$
$$n=10\rightarrow =100\times 121=\boxed{12100}$$
The value of $$\sum^{10}_{x=1}\sum^{r=x-1}_{r=0}(2^{x}-2^{r})$$ is
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$$16392$$
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$$16398$$
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$$15462$$
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$$15468$$
Explanation
$$\displaystyle\sum_{x=1}^{10}\sum_{r=0}^{x-1}(2^{x}-2^{r})$$
Now, $$\displaystyle\sum_{r=0}^{x-1}(2^{x}-2^{r})=(2^{x}-2^{1})+(2^{x}-2^{2})+......+(2^{x}-2^{x-1})$$
$$=(2^{x}+2^{x}+2^{x}+........+x\ terms)-(2^{0}+2^{1}+2^{2}+......+2^{x-1})$$
$$=x.2^{x}-\left[\dfrac{2^{0}(2^{x}-1)}{2-1}\right]=x.2^{x}-(2^{x}-1)=x.2^{x}-2^{x}+1$$
Now, $$\displaystyle\sum_{x=1}^{10}\sum_{r=0}^{x-1}(2^{c}-2^{r})=\sum_{x=1}^{10}x.2^{x}+\sum_{x=1}^{10}1$$
$$=\displaystyle\sum_{x=1}^{10}x.2^{x}-\left[2^{1}+2^{2}+2^{3}+........+2^{10}\right]+10$$
$$=\displaystyle\sum_{x=1}^{10}x.2^{x}-\left[\dfrac{2(2^{10}-1)}{2-1}\right]+10=\sum_{x=1}^{10}x.2^{x}-2^{11}+12$$
$$=(1.2+2.2^{2}+3.2^{3}+......+10.2^{10})-2^{11}+12$$
$$=18734-2048+12=\boxed{16398}$$ Ans
Observe the pattern carefully
$$11\times11=121$$
$$111\times111=12321$$
$$1111\times1111=\,?$$
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$$12345321$$
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$$123421$$
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$$1234321$$
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$$12468$$
Explanation
$$11\times11=121$$
$$111\times111=12321$$
$$1111\times1111=1234321$$
Sum of the series
$$(1\times 2015)+(2 \times 2014)+(3\times 2013)......+(2015\times 1)$$ is equal to-
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$$336\ \times 2015 \times 2016$$
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$$336\ \times 2015 \times 2017$$
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$$336\ \times 2016\times 201$$
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$$None$$
It is known that $$\sum\limits_{r = 1}^\infty {\frac{1}{{{{(2r - 1)}^2}}} = \frac{{{\pi ^2}}}{8},} $$ then $$\sum\limits_{r = 1}^\infty {\frac{1}{{{r^2}}}} $$ is equal to
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$$\frac{{{\pi ^2}}}{{24}}$$
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$$\frac{{{\pi ^2}}}{{3}}$$
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$$\frac{{{\pi ^2}}}{{6}}$$
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none of these
Explanation
Given : $$\displaystyle \sum_{r = 1}^{\infty } \frac{1}{(2r-1)^{2}} = \frac{\pi ^{2}}{8}, \sum_{r = 1}^{\infty } \frac{1}{r^{2}} = ?? $$
start putting 'r' value
$$\displaystyle \frac{1}{1^{2}}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}------ = \frac{\pi ^{2}}{8} $$
let say
$$\displaystyle \sum_{r = 1}^{\infty } \frac{1}{r^{2}} = k $$
$$\displaystyle \frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+-----\frac{1}{r^{2}} = k $$
$$\displaystyle [\frac{1}{1^{2}}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+-----\frac{1}{(2r-1)^{2}}]+\frac{1}{2^{2}}+\frac{1}{4^{2}}+---- = k $$
$$\displaystyle \frac{\pi ^{2}}{8}+\frac{1}{2^{2}}[\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}----] = k $$
$$\displaystyle \frac{\pi ^{2}}{8}+\frac{1}{4}(k) = k $$
$$\displaystyle \frac{\pi ^{2}}{8} = k-\frac{k}{4} = \frac{3k}{4} $$
so, $$\displaystyle k = \frac{\pi ^{2}}{6} $$ Answer option- C
Find the sum of the infinite series $$\frac{1}{9}+\frac{1}{81}+\frac{1}{729}+....\infty$$
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$$\frac{1}{8}$$
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$$\frac{1}{4}$$
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$$\frac{1}{5}$$
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$$\frac{2}{3}$$
The sum $$\frac{3}{{1.2}},\frac{3}{{1.2}},\frac{1}{2},\frac{4}{{2.3}}{\left( {\frac{1}{2}} \right)^2} + \frac{5}{{3.4}}{\left( {\frac{1}{2}} \right)^2}$$
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$$1 - \frac{1}{{\left( {n + 1} \right){2^n}}}$$
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$$1 - \frac{1}{{n{{.2}^{n - 1}}}}$$
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$$1 = \frac{1}{{\left( {n + 1} \right){2^n}}}$$
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$$\frac{1}{{\left( {n - 1} \right){2^{n - 1}}}}$$
Explanation
$$t_n=\dfrac{n+2}{n(n+1)}\cdot \left(\dfrac{1}{2}\right)^n$$
$$=\dfrac{2(n+1)-n}{n(n+1)} \cdot \left(\dfrac{1}{2}\right)^n$$
$$t_n=\dfrac{1}{n}\left(\dfrac{1}{2}\right)^{n-1}-\dfrac{1}{n+1} \cdot \left(\dfrac{1}{2}\right)^n$$
$$S_n=\sum_{n=1}^nt_n$$
$$=\left\{\dfrac{1}{1}\left(\dfrac{1}{2}\right)^0-\dfrac{1}{2}\left(\dfrac{1}{2}\right)^1\right\}$$
$$+\left\{\dfrac{1}{2}\left(\dfrac{1}{2}\right)^1-\dfrac{1}{3}\left(\dfrac{1}{2}\right)^2\right\}+...+$$
$$=\left\{\dfrac{1}{n}\left(\dfrac{1}{2}\right)^{n-1}-\dfrac{1}{n+1}\left(\dfrac{1}{2}\right)^n\right\}$$
$$S_n=1-\dfrac{1}{(n+1)2^n}$$
If f(x)= $$x+\frac{1}{2x+\frac{1}{\frac{1}{2x+...\infty }}}$$ ;
then the value of f (2011). f'(2011) is :
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0
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1
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2011
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2010
$$1 + \frac { n } { 2 } + \frac { n ( n - 1 ) } { 2.4 } + \frac { n ( n - 1 ) ( n - 2 ) } { 2.4 .6 } + \dots \ldots \ldots \infty =$$ ?
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$$\left( \frac { 2 } { 3 } \right) ^ { n }$$
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$$\left( \frac { 3 } { 2 } \right) ^ { n }$$
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$$\left( \frac { 3 } { 4 } \right) ^ { n }$$
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$$\left( \frac { 4 } { 3 } \right) ^ { n }$$
Find the missing number, if same rule is followed in all the three figures.
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12
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16
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14
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9
Explanation
REF.Image.
Given
If we consider down two number of fig(i)
then their product is $$ \Rightarrow 4\times 16 = 64 $$
Now if we take square root then =$$ \sqrt{64}$$
$$ = 8 \to$$upper are we are getting.
similarly from fig (iii)
down number product $$ \Rightarrow 12 \times27 = 324$$
take square roots $$ = \sqrt{324} = 18 \to$$upper number.
So, similarly from fig(ii)
Down product = $$ 18 \times 8 = 144 $$
so square root = $$ \sqrt{144} = 12 \rightarrow $$ option A
Choose the
CORRECT
options:-
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$$\sum\limits_{k = 1}^{360} {\left( {\frac{1}{{k\sqrt {k + 1} + \left( {k + 1} \right)\sqrt k }}} \right)} $$ is the ratio of two relative prime positive integers $$m$$ nad $$n$$. The value of $$(m+n)$$ is equal to $$37$$.
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If $$\frac{{5\pi }}{2} < x < 3\pi $$, then the value of the expression $$\frac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}$$ is $$ - \tan \frac{x}{2}$$.
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The exact value of $$\frac{{96\sin 80^\circ \sin 65^\circ \sin 35^\circ }}{{\sin 20^\circ + \sin 50^\circ + \sin 110^\circ }}$$ is equal to $$24$$.
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The sum $$\sum\limits_{n = 1}^\infty {\left( {\frac{n}{{{n^4} + 4}}} \right)} $$ is equal to $$3/16$$.
$$\sum\limits_{r = 1}^n {{{\sin }^{ - 1}}\left( {\frac{{\sqrt r - \sqrt {r - 1} }}{{\sqrt {r\left( {r + 1} \right)} }}} \right)} $$ is equal to
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$${\tan ^{ - 1}}\left( {\sqrt n } \right) - \frac{\pi }{4}$$
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$${\tan ^{ - 1}}\left( {\sqrt n + 1} \right) - \frac{\pi }{4}$$
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$${\tan ^{ - 1}}\left( {\sqrt n } \right)$$
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$${\tan ^{ - 1}}\left( {\sqrt n + 1} \right)$$
Direction: In a given question, one number is missing in the series. you have to understand the pattern of the series and insert the number
JN 28 27 GP
CE 12 45 TU
LR ? ? MS
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$$34, 36$$
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$$35, 35$$
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$$30, 32$$
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$$30, 41$$
Let $$A$$ be the sum of the first $$20$$ terms and $$B$$ be the sum of the first $$40$$ terms of the series $$1$$ + $$2.2^2$$ + $$3^2$$ + $$2.4^2$$ + $$5^2+2.6^2$$ +......... Find the value of $$A$$.
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$$496$$
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$$232$$
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$$248$$
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$$464$$
Explanation
Sum of first $$20$$ terms -
$$\Rightarrow \left( { 1 }^{ 2 }+{ 3 }^{ 2 }+\_ \_ \_ \_ +{ 19 }^{ 2 } \right) +2\left( { 2 }^{ 2 }+{ 4 }^{ 2 }+\_ \_ \_ \_ +{ 20 }^{ 2 } \right) $$
$$\Rightarrow A=\left( { 1 }^{ 2 }+{ 2 }^{ 2 }+3^2\_ \_ \_ \_ +{ 19 }^{ 2 }+20^2 \right) +\left( { 2 }^{ 2 }+{ 4 }^{ 2 }+\_ \_ \_ \_ +{ 20 }^{ 2 } \right) $$
$$\Rightarrow A=\dfrac{20\times 41\times 21}{6}+1^2\left( { 1 }^{ 2 }+{ 2 }^{ 2 }+\_ \_ \_ \_ +{ 10 }^{ 2 } \right) $$
$$\Rightarrow A=2870+4\times \dfrac{10\times 11\times 21}{6}$$
$$\Rightarrow A=2870+1540$$
$$\Rightarrow A=4410.$$
Hence, the answer is $$4410.$$
If $$1^{2}+2^{2}+3^{2}++(2003) ^{2}=(2003)(4007)(334)$$ and $$(1)(2003)+ (2)(2002)= (2003)(334)(x)$$ then $$x$$ equals
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$$2005$$
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$$2004$$
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$$2003$$
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$$2001$$
The number of the three digits numbers having only two consecutive digits identical is:
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153
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160
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180
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161
The sum of the first n terms of the series $$\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+$$ is equal to:
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$$2^{-n}- n -1$$
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$$1 - 2^{-n}$$
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$$n + 2^{-n} - 1$$
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$$2^{-n} - 1$$
$$1+(1+2)+(1+2+{2}^{2})+(1+2+{2}^{2}+{2}^{3})=....$$ upto $$n$$ terms $$=$$______
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$${2}^{n+2}-n-4$$
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$$2({2}^{n}-1)-n$$
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$${2}^{n+1}-n$$
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$${2}^{2n+1}-1$$
If $$24 + 37 = 7,12 + 18 = 3$$, then 54+21=?
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11
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9
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12
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3
In the following number series,one number is wrong.find out the ?
1,2,8,33,149,765,4626
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$$33$$
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$$8$$
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$$149$$
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$$0$$
Explanation
1,2,8,33,149,765,4626
Carefully analyzing the series, we can see the following pattern
$$1\times 1+(1)^{2}=2$$
$$2\times 2+(2)^{2}=8$$
$$8\times 3+(3)^{2}=33$$
$$33\times 4+(4)^{2}=148$$
$$148\times 5+(5)^{2}=765$$
$$765\times 6+(6)^{2}=4626$$
So, we can clearly see that 149 is the wrong entry in the series.
If $$ '-'$$ denotes $$'\div ',\ '\div '$$ denotes $$'\times ',\ '+'$$ denotes' - 'and $$'\times '$$ denotes $$'+'$$, then find the value of $$116+9\div 52-4\times 5$$.
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$$16$$
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$$8$$
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$$9$$
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$$4$$
Explanation
$$-$$ denoted $$\div$$
$$\div$$ denoted $$\times $$
$$+$$ denoted $$-$$
$$\times $$ denoted $$+$$ $$\boxed{BODMAS}$$
$$116+9\div 52-4\times 5$$
$$116-9\times 52\div 4+5$$
$$116-9\times 13+5$$
$$116-117+5$$
$$116-112=4$$
$$1+\dfrac{1}{2}(1+2)+\dfrac{1}{3}(1+2+3)+\dfrac{1}{4}(1+2+3+4)+.....$$ upto $$20$$ terms is
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$$110$$
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$$111$$
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$$115$$
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$$116$$
Explanation
Given
$$S=1+\cfrac{1}{2}(1+2)+\cfrac{1}{3}(1+2+3)+.....+\cfrac{1}{20}(1+2+3+....+20)$$
$$\displaystyle=\sum _{ n=1 }^{ 20 }{ \cfrac { 1 }{ n } } \sum { n } $$
$$\displaystyle \sum _{ n=1 }^{ 20 }{ \cfrac { 1 }{ n } } .\cfrac{n(n+1)}{2}=\cfrac{1}{2}\sum _{ n=1 }^{ 20 }{ (n+1) } $$
$$\cfrac { 1 }{ 2 } { \left[ \cfrac { n(n+1) }{ 2 } +n \right] }_{ 0 }^{ 20 }$$
$$=\cfrac{1}{2}\left[ \cfrac { 20(21) }{ 2 } +20 \right] $$
$$=5(21)+10=115$$
Find the missing term in the series given below. $$12,\ 13,\ 18,\ 19,\ 24,\ 25 $$?
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$$27$$
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$$30$$
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$$29$$
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$$26$$
Explanation
$$12,13,18,19,24,25$$
difference between $$12,13=1$$
difference between $$13,18=5$$
Similarly
difference between $$18,19=1$$
difference between $$19,24=5$$
similarly
difference between $$24,25=1$$
so to get next number of $$25$$ add $$5$$ i.e., $$25+5=30$$
Let a sequence whose $$n^{th}$$ term is $${a_n}$$ be define as $${a_1} = \frac{1}{2}$$ and $$(n - 1){a_{n - 1}} = (n + 1){a_n}$$ for $$n \ge 2$$ ; then
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$${a_n} = \frac{1}{{n(n + 1)}}$$
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$${S_n} = 1 - \frac{1}{{n + 1}}$$
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$$\mathop {\lim }\limits_{n \to \infty } {S_n} = 1$$
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$${S_n} = \frac{1}{{n + 1}}$$
Explanation
$$a_{1}=\cfrac{1}{2}$$
$$(n+1)a_{n}=(n-1)a_{n-1}$$, $$n \ge 2$$
For,
$$n=2$$, $$3a_{2}=a_{1}$$
$$n=3$$, $$4a_{3}=2a_{2}$$
$$n=4$$, $$5a_{4}=3a_{3}$$
.................................
$$n=n$$, $$(n+1)a_{n}=(n-1)a_{n-1}$$
Multiplying the above, we get
$$3a_{2}\times 4a_{3}\times 5a_{4}...........\times(n+1)a_{n}=a_{1}\times 2a_{2}\times 3a_{3}\times 4a_{4}\times ..............\times (n-1)a_{n-1}$$
$$\Rightarrow 3a_{2}\times 4a_{3}\times 5a_{4}...........\times(n+1)a_{n}=\cfrac{1}{2}\times 2a_{2}\times 3a_{3}\times 4a_{4}\times ..............\times (n-1)a_{n-1}$$
$$\Rightarrow 3a_{2}\times 4a_{3}\times 5a_{4}...........\times(n)a_{n-1}(n+1)a_{n}=\cfrac{1}{2}\times 2a_{2}\times 3a_{3}\times 4a_{4}\times ..............\times (n-1)a_{n-1}$$
$$\Rightarrow n(n+1)a_{n}=1$$
$$\Rightarrow a_{n}=\cfrac{1}{n(n+1)}=\cfrac{1}{n}-\cfrac{1}{n+1}$$
$$\therefore S_{n}=\sum a_{n}=\sum_{n=1}^{n}[\cfrac{1}{n}-\cfrac{1}{n+1}]=[\cfrac{1}{1}-\cfrac{1}{2}+\cfrac{1}{2}-\cfrac{1}{3}+\cfrac{1}{3}-\cfrac{1}{4}.......\cfrac{1}{n}-\cfrac{1}{n+1}]$$
$$\Rightarrow S_{n}=1-\cfrac{1}{n+1}$$
Now, $$\lim_{n\rightarrow \infty}S_{n}=\lim_{n\rightarrow \infty}1-\cfrac{1}{n+1}=1-\lim_{n\rightarrow \infty}\cfrac{1}{n+1}=1$$
The sum of the series 1.2.3+2.3.4+...3.4.5.+...to n terms is
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$$n(n+1)(n+2)$$
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$$(n+1)(n+2)(n+3)$$
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$$\frac { 1 }{ 4 } n(n+1)(n+2)(n+3)$$
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$$\frac { 1 }{ 4 } (n+1)(n+2)(n+3)$$
The sum of first $$9$$ terms of the series
$$\dfrac { { 1 }^{ 3 } }{ 1 } +\dfrac { { 1 }^{ 3 }+{ 2 }^{ 3 } }{ 1+3 } +\dfrac { { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 } }{ 1+3+5 } +\dots$$
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$$96$$
0%
$$142$$
0%
$$192$$
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$$71$$
If $$\left( { 1-y } \right) ^{ 35 }\left( 1+y \right) ^{ 45 }={ A }_{ 0 }+{ A }_{ 1 }Y+{ A }_{ 2 }{ Y }^{ 2 }+{ A }_{ 3 }{ Y }^{ 3 }+.....+{ A }_{ 80 }{ Y }^{ 80 },$$ then
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$$\frac { { A }_{ 2 } }{ A_{ 1 } } <2$$
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$${ A }_{ 1 }={ A }_{ 2 }$$
0%
$$\frac { { A }_{ 2 } }{ { A }_{ 1 } } <1$$
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$$1<\frac { { A }_{ 2 } }{ { A }_{ 1 } } <2$$
If $$\dfrac{1}{1^{2}}+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+........\infty =\dfrac{\pi^{2}}{6}$$ then $$\dfrac{1}{1^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{5^{2}}+......\infty$$
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0%
$$\dfrac{\pi^{2}}{8}$$
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$$\dfrac{\pi^{2}}{12}$$
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$$\dfrac{\pi^{2}}{3}$$
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$$\dfrac{\pi^{2}}{2}$$
Sum of the series
$$1+2.2+3.2^{2}+...+100.2^{10}=$$
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$$100.2^{100}+1$$
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$$99.2^{100}+1$$
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$$99.2^{100}-1$$
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$$100.2^{100}-1$$
Sum of series $$\displaystyle \sum^{n}_{r=1}(r^{2}+1)_{r!}$$ is
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$$(n+1)!$$
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$$(n+2)!-1$$
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$$n(n+2)!$$
0%
$$(n+1)!-1$$
$$\dfrac {1}{1.4}+\dfrac {1}{4.7}+\dfrac {1}{7.10}+...+\dfrac {1}{(3n-5)(3n-2)}$$
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0%
$$\dfrac {n-1}{3n+2}$$
0%
$$\dfrac {n-1}{3n-2}$$
0%
$$\dfrac {n-1}{3n-1}$$
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$$\dfrac {n-1}{3n-4}$$
If $$x=\sqrt 1+\dfrac{1}{{1}^{2}}+\dfrac{1}{{2}^{2}}+\sqrt 1+\dfrac{1}{{2}^{2}}+\dfrac{1}{{3}^{2}}+\sqrt 1+\dfrac{1}{{2019}^{2}}+\dfrac{1}{{2020}^{2}}$$ then $$4040(2020-x)$$
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$$1$$
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$$2$$
0%
$$-1$$
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$$-2$$
The sum of the series $$1+\frac { 1.3 }{ 6 } +\frac { 1.3.5 }{ 6.8 } +...\infty $$ is
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0%
$$4$$
0%
$$0$$
0%
$$\infty $$
0%
$${ (4) }^{ \frac { 2 }{ 3 } }$$
The sum of the infinite series $$\cot { ^{ 1 } } \left( \dfrac { 7 }{ 4 } \right) +\cot { ^{ -1 }\left( \dfrac { 19 }{ 4 } \right) } +\cot { ^{ -1 }\left( \dfrac { 39 }{ 4 } \right) } +\cot { ^{ -1 }\left( \dfrac { 67 }{ 4 } \right) } +........\infty $$ is:
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$$\\ \dfrac { \pi }{ 4 } -\cot { ^{ -1 } } (3)$$
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$$\\ \dfrac { \pi }{ 4 } -\tan { ^{ -1 } } (3)$$
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$$\\ \dfrac { \pi }{ 4 } +\cot { ^{ -1 } } (3)$$
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$$\\ \dfrac { \pi }{ 4 } +\tan { ^{ -1 } } (3)$$
$$40280625, 732375, 16275, 465, 18.6, 1.24,?$$
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0%
$$0.248$$
0%
$$0.336$$
0%
$$0.424$$
0%
$$0.512$$
0%
$$0.639$$
Explanation
$$40280625, 732375,16275,465,18.6,1.24$$
$$\cfrac{40280625}{732375}=55$$
$$\cfrac{732375}{16275}=45$$
$$\cfrac{16275}{465}=35$$
$$\cfrac{465}{18.6}=25$$
$$\cfrac{18.6}{1.24}=15$$
$$\cfrac{1.24}{x}=5$$
$$x=0.248$$
Sum of the series $$ S = 1^{2}-2^{2}+3^{2}-4^{2}+......-2002^{2}+2003^{2}$$ is
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0%
2007006
0%
1005004
0%
2000506
0%
none of these
In the sum $$3+33+333+3333+.........2015$$ terms the number formed by taking the last four digits in that order is
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$$6365$$
0%
$$6255$$
0%
$$6465$$
0%
$$6565$$
Sum of the series $$\dfrac{1^2}{1!}+\dfrac{2^2}{2!}+\dfrac{3^2}{3!}+\dfrac{4^2}{4!}+.....$$(infinite terms) is?
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0%
e
0%
$$2e$$
0%
$$e^2$$
0%
$$\infty$$
Explanation
$$ \begin{aligned} &S=\frac{1^{2}}{1 !}+\frac{2^{2}}{2 !}+\frac{3^{2}}{3 !}+\frac{4^{2}}{4 b}+\\ &\text {(infinite terems)} is \\ &S=\sum_{n=1}^{\infty} n^{2}\\ &n^{2}=a_{0}+a_{1} n+a_{2} n(n-1)\\ &n=0 \quad 0=a_{0}\\ &n=1 \quad 1=0+a, \Rightarrow a_{1}=1\\ &n=2 \quad 4=0+2+2 a_{2} \Rightarrow a_{2}=1\\ &S=\sum_{n=1}^{\infty}\left(a_{0}+a_{1} n+a_{2} n(n-1)\right) \end{aligned} $$
$$S=\sum_{n=1}^{\infty}\left(\frac{0+n+n(n-1)}{n_{0}}\right)$$ $$S=\sum_{n=1}^{\infty} \frac{n}{n_{b}}+\sum_{n=1}^{\infty} \frac{n(n-1)}{n_{0}^{1}}$$
$$S=e+e=2 e$$
$$e=1+\frac{1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\cdots} \Rightarrow \sum_{n=1}^{\infty} \frac{1}{(n-1) !}$$ $$e=\sum_{n=2(n-2)\}}=1+1+\frac{1}{2 !}+\frac{1}{3 !} \cdots \cdot \cdots$$
Option $$B$$ is correct
If $$a_{k}=\dfrac{1}{K(K+1)}$$ for $$K=1, 2, 3,.....n,$$ then $$\left(\sum_{k=1}^{n}{a_{k}}\right)^{2}=$$
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0%
$$\dfrac{n}{n+1}$$
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$$\dfrac{n^{2}}{(n+1)^{2}}$$
0%
$$\dfrac{n^{4}}{(n+1)^{4}}$$
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$$\dfrac{n^{6}}{(n+1)^{6}}$$
Explanation
$$\begin{array}{l} { a_{ k } }=\dfrac { 1 }{ { k\left( { k+1 } \right) } } =\dfrac { 1 }{ k } -\dfrac { 1 }{ { k+1 } } \\ { \left( { \sum _{ k=1 }^{ n }{ { a_{ x } } } } \right) ^{ 2 } } \\ { \left( { \sum _{ k=1 }^{ n }{ \left( { \dfrac { 1 }{ k } -\dfrac { 1 }{ { k+1 } } } \right) } } \right) ^{ 2 } } \\ =\left( { 1-\dfrac { 1 }{ 2 } } \right) +\left( { \dfrac { 1 }{ 2 } -\dfrac { 1 }{ 3 } } \right) \, \, \, \, \, \left( { \dfrac { 1 }{ n } \dfrac { { -1 } }{ { n+1 } } } \right) \\ ={ \left( { 1-\dfrac { 1 }{ { n+1 } } } \right) ^{ 2 } } \\ ={ \left( { \dfrac { { n+1-1 } }{ { n+1 } } } \right) ^{ 2 } } \\ =\dfrac { { { n^{ 2 } } } }{ { { { \left( { n+1 } \right) }^{ 2 } } } } \end{array}$$
Hence, this is the answer.
$$\dfrac { { C }_{ 1 } }{ { C }_{ 0 } } +2\dfrac { { C }_{ 2 } }{ { C }_{ 1 } } +3\frac { { C }_{ 3 } }{ { C }_{ 2 } } +......+\dfrac { n{ C }_{ n } }{ { C }_{ n-1 } } $$ equals
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0%
$${ n2 }^{ n-1 }$$
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$$\dfrac { { n2 }^{ n-1 } }{ { 2 }^{ n }-1 } $$
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$$\dfrac { n(n+1) }{ 2 } $$
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$$\dfrac { (n+1)(n+2) }{ 2 } $$
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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