CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 4 - MCQExams.com

We would like to regroup the series as follows:
$$ 1^2+(3^2-2^2)+(5^2-4^2)+....(1001^2-1000^2) $$
$$a^2-b^2=(a+b)(a-b)$$ , in our case $$a-b=1$$
Hence,
$$1+5+9+13+... +2001$$
There are $$501$$ terms, then
Sum$$=501\times1001$$

Numerator : $$\sum_{k=1}^n k(k+1)^2$$ $$ = \sum_{k=1}^n (k^3+2k^2+k) $$
Denominator : $$\sum_{k=1}^n k^2(k+1)$$ = $$ \sum_{k=1}^n (k^3 + k^2) $$
On evaluating each sum we get
Numerator = $$n(n+1)(\frac{n(n+1)}{4} + \frac{2n+1}{3} + \frac{1}{2}) $$
Denominator = $$n(n+1)(\frac{n(n+1)}{4}+\frac{2n+1}{6}) $$
Now,
$$Answer=\frac{numerator}{denominator}=\frac{n(n+1)}{n(n+1)}\times\frac{(n+2)(3n+5)}{(n+2)(3n+1)}=\frac{3n+5}{3n+1}$$
Hence, option 'B' is correct.

$$S=\Sigma n^2 + \Sigma n $$
Therefore,
$$S=\dfrac{n(n+1)(2n+1)}6+\dfrac{n(n+1)}2 $$
Hence,
$$S=\dfrac{n(n+1)(n+2)}3 $$

We can see that first 6 terms taken in pair and is equal to  $$-3+(-7)+(-11)$$
So the sum will be $$-21+7^2$$
$$=28$$

$$S=\Sigma r^3 +\Sigma r^2 +\Sigma r$$

$$\displaystyle \mathrm{V}_{\mathrm{r}}=\frac{\mathrm{r}}{2}[2\mathrm{r}+(\mathrm{r}-1)(2\mathrm{r}-1)]=\frac{1}{2}(2\mathrm{r}^{3}-\mathrm{r}^{2}+\mathrm{r}) $$
$$\therefore \displaystyle \sum \mathrm{V}_{\mathrm{r}}=\frac{1}{2}\left(2 \sum r^3-\sum r^2+\sum r\right)=\frac{1}{12}\mathrm{n}(\mathrm{n}+1)(3\mathrm{n}^{2}+\mathrm{n}+2)$$ .

$$\therefore t_{r}=\dfrac {\Sigma k^2}{\Sigma k}=\dfrac {\dfrac {n(n+1)(2n+1)}{6}}{\dfrac {n(n+1)}{2}}=\dfrac {2n+1}{3}$$

General term is given by $$(2k)(1+3k)$$ where $$k$$ varies from $$1$$  to $$n-1$$
$$ \displaystyle  \sum_{k=1}^{n-1} (2k)(1+3k) = 2 \sum_{k=1}^{n-1} (k+3k^2)$$.
Substituting the values in the formula (Note that $$n = n-1$$),
We get $$(n)(n-1)(1+(2(n-1)+1))$$$$=2n^3-2n^2$$

For odd, let $$n=2m+1$$
$$\Rightarrow { S }_{ 2m+1 }={ S }_{ 2m }+{ (2m+1) }^{ th } term$$      $$...(1)$$
Also, given that $${ S }_{ 2m }=\dfrac { m{ (m+1) }^{ 2 } }{ 2 } $$.
Since, $$n=2m+1$$
$$\Rightarrow 2m=n-1$$
So, $${ S }_{ 2m }=\dfrac { (n-1){ (n-1+1) }^{ 2 } }{ 2 } $$
$$\Rightarrow { S }_{ 2m }=\dfrac { (n-1){ n }^{ 2 } }{ 2 } $$      $$...(2)$$
Also, the odd terms of the given series are $${ 3 }^{ 2 },{ 5 }^{ 2 },...etc$$.
$$\Rightarrow { (2m+1) }^{ th } term={ (2m+1) }^{ 2 }$$
$$\Rightarrow { (2m+1) }^{ th } term={ n }^{ 2 }$$      $$...(3)$$
Now, substitute $$(2)$$ and $$(3)$$ in $$(1)$$; we get
$${ S }_{ 2m+1 }=\dfrac { (n-1){ n }^{ 2 } }{ 2 } +{ n }^{ 2 }$$

$$ x=(-1)^{ { a }^{ 1 } }+(-1)^{ { a }^{ 2 } }+..............+(-1)^{ { a }^{ 1006 } }\\ \because \quad a\quad is\quad not\quad distinct\quad and\quad it\quad is\quad an\quad integer.\\ Let\quad us\quad take\quad a\quad as\quad even.\\ \therefore \quad { a }^{ n }\quad will\quad be\quad even\quad for\quad all\quad values\quad of\quad n.\\ \therefore \quad (-1)^{ { a }^{ n } }\quad will\quad be\quad 1\\ Total\quad number\quad of\quad terms\quad is\quad 1006.\\ \therefore \quad x=\sum _{ n=1 }^{ 1006 }{ (-1)^{ n } } =1006\\ \therefore \quad x=1006\longrightarrow an\quad even\quad number.\\ Again,\quad let\quad a\quad be\quad odd.\quad Then\quad { a }^{ n }\quad will\quad be\quad odd\quad for\quad all\quad \\ values\quad of\quad n.\\ \therefore \quad (-1)^{ { a }^{ n } }=-1\\ \therefore \quad \sum _{ n=1 }^{ 1006 }{ (-1)^{ n } } =\quad -1006\rightarrow an\quad even\quad negative\quad number.\\ Again\quad y=(-1)^{ { a }^{ 1007 } }+(-1)^{ { a }^{ 1008 } }+...................+(-1)^{ { a }^{ 2013 } }\\ As\quad per\quad the\quad previous\quad discussion\\ (-1)^{ { a }^{ n } }=1\quad in\quad case\quad a\quad is\quad even\quad and\quad \\ (-1)^{ { a }^{ n } }=\quad -1\quad in\quad case\quad a\quad is\quad odd\\ Total\quad number\quad of\quad terms=(2013-1007)+1=1007\\ Then\quad y=1007\quad in\quad case\quad a\quad is\quad even\\ \quad \quad \quad \quad \quad =-1007\quad in\quad case\quad a\quad is\quad odd\\ y\quad is\quad an\quad odd\quad number,\quad positive\quad and\quad negative\quad respectively.\\ Let\quad us\quad investigate\quad the\quad options.\\ In\quad case\quad a\quad is\quad odd\longrightarrow \left\{ x\quad is\quad even,\quad negative\\ y\quad is\quad odd,\quad negative \right\} ----(1)\\ In\quad case\quad a\quad is\quad even\longrightarrow \left\{ x\quad is\quad even,\quad positive\\ y\quad is\quad odd,\quad positive \right\} ----(2)                                                                      Option\quad A\longrightarrow (-1)^{ x }=1,\quad (-1)^{ y }=-1\longrightarrow a\quad odd.\\ a\quad even\longrightarrow (-1)^{ x }=1,\quad (-1)^{ y }=-1\\ Option\quad A\quad is\quad not\quad correct.\\ Option\quad B\longrightarrow a\quad is\quad odd\longrightarrow (-1)^{ x }=1,\quad (-1)^{ y }=-1\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad a\quad is\quad even\longrightarrow (-1)^{ x }=1,\quad (-1)^{ y }=-1\\ Option\quad B\quad is\quad correct.\\ Obviously\quad option\quad C\quad and\quad D\quad are\quad not\quad correct.\\ Answer-\quad Option\quad B\quad is\quad correct. $$

After each term starting with $$p_3$$, the number in from of 4 is twice the number in front of 4 in the preceding term.
Thus the number for the unknown plant is $$2 \times 12 +4 =28$$

Ratio between first two terms $$=$$ $$\dfrac{30}{15}$$ $$=2$$
Ratio between second and third terms $$=$$ $$\dfrac{60}{30}$$ $$=2$$
Ratio between third and fourth terms $$=$$ $$\dfrac{120}{60}$$ $$=2$$
Ratio between fourth and fifth terms $$=$$ $$\dfrac{240}{120}$$ $$=2$$
Since, the ratio between the terms is the same. The series forms a G.P.

$$\displaystyle a_n$$ is sum of reciprocals of natural numbers starting from $$n + 1$$ and ending at $$3n$$.
$$\displaystyle \therefore a_2=\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6}=\frac {19}{20}$$
$$\displaystyle \therefore a_{n+1}=\frac {1}{n+2}+\frac {1}{n+3}+ ....+\frac {1}{3n}+\frac {1}{3n+1}+\frac {1}{3n+2}+\frac {1}{3n+3}$$
$$\displaystyle \Rightarrow a_{n+1}-a_n=\frac {1}{3n+1}+\frac {1}{3n+2}+\frac {1}{3n+3}-\frac {1}{n+1}$$
$$\displaystyle =\frac {1}{3n+1}+\frac {1}{3n+2}-\frac {2}{3n+3}$$
$$\displaystyle =\frac {(3n+2)(3n+3)}{(3n+1)(3n+2)(3n+3)}+\frac {(3n+1)(3n+3)}{(3n+1)(3n+2)(3n+3)}-\frac {2(3n+1)(3n+2)}{(3n+1)(3n+2)(3n+3)}$$
$$\displaystyle =\frac {9n+5}{(3n+1)(3n+2)(3n+3)}$$
Hence, options 'B' and 'C' are correct.

From the sequence give we can find general formula
$$a_n = a_{n-2}+ a_{n-1}$$
$$a_8 = a_7+a_6 =13 +8=21$$
$$a_9 = a_8+a_7 = 21+13 = 34$$
$$a_{10}= a_9+a_8 = 34+21=55$$
$$a_{11} = a_{10}+a_9 = 55 + 34 = 89$$
$$a_{12} = a_{10}+ a_{11}=55+89=144$$

Part of base will be 8 times the part of red pigment.

Parts of red pigment	1	4	7	12	20
Parts of base	8	32	56	96	160

$$a_1=1, a_{n+1}^{2}=(2\alpha^2-2\alpha+1)^n$$

The pattern of numbers are as followed
$$1^2-1=0$$
$$2^2-1=3$$
$$3^2-3=6$$
$$4^2-4=12$$
$$5^2-5=20$$
$$6^2-6=30$$
$$7^2-7=42$$
$$8^2-8=56$$
$$9^2-9=72$$

In series $$2,4,6,8,....$$ difference is same i.e. $$2$$

$$\displaystyle \sum_{n=1}^{10} \int_{0}^{n} x dx =  \sum_{n=1}^{10} \frac{n^{2}}{2} $$

The sequence can be written as :
$${ S }_{ n }={ n }^{ 3 }+{ \left( n-2 \right)  }^{ 3 }+{ \left( n-4 \right)  }^{ 3 }+......+{ 1 }^{ 3 }-\left[ { \left( n-1 \right)  }^{ 3 }+{ \left( n-3 \right)  }^{ 3 }+{ \left( n-5 \right)  }^{ 3 }+....+{ 2 }^{ 3 } \right] \\ ={ n }^{ 3 }+{ \left( n-2 \right)  }^{ 3 }+{ \left( n-4 \right)  }^{ 3 }+......+{ 1 }^{ 3 }+\left[ { \left( n-1 \right)  }^{ 3 }+{ \left( n-3 \right)  }^{ 3 }+{ \left( n-5 \right)  }^{ 3 }+....+{ 2 }^{ 3 } \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -2\left[ { \left( n-1 \right)  }^{ 3 }+{ \left( n-3 \right)  }^{ 3 }+{ \left( n-5 \right)  }^{ 3 }+....+{ 2 }^{ 3 } \right] \\ =\sum _{ r=1 }^{ n }{ r^{ 3 } } -2\times { 2 }^{ 3 }\left[ { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+............+{ \left( \dfrac { n-1 }{ 2 }  \right)  }^{ 3 } \right] $$
we have taken 8 common from the 2nd part of the sequence.
$$=\sum _{ r=1 }^{ n }{ r^{ 3 } } -16\sum _{ 1 }^{ \dfrac { n-1 }{ 2 }  }{ r^{ 3 } } \\ =\left[ { \dfrac { n\left( n+1 \right)  }{ 2 }  }^{ 2 } \right] -16\left[ { \dfrac { \dfrac { n-1 }{ 2 } \left( \dfrac { n-1 }{ 2 } +1 \right)  }{ 2 }  }^{ 2 } \right] \\ ={ \dfrac { n\left( n+1 \right)  }{ 2 }  }^{ 2 }-16\left[ { \dfrac { \left( n-1 \right) \left( n+1 \right)  }{ 8 }  }^{ 2 } \right] \\ =\dfrac { { \left( n+1 \right)  }^{ 2 } }{ 4 } \left[ { n }^{ 2 }-{ \left( n-1 \right)  }^{ 2 } \right] \\ =\dfrac { { \left( n+1 \right)  }^{ 2 }\left( 2n-1 \right)  }{ 4 } $$

If $$n$$ is even,
Let $$n=2m$$
$$\therefore S=1^{3}+3.2^{2}+3^{3}+3.4^{2}+5^{3}+3.6^{2}+....+(2m-1)^{3}+3(2m)^{2}$$

$$=\left \{ 1^{3}+3^{3}+5^{3}+...+(2m-1)^{3} \right \}+3\left \{ 2^{2}+4^{2}+6^{2}+....+(2m)^{2} \right \}$$

$$=\sum_{r=1}^{m}(2r-1)^{3}+3.4\sum_{r=1}^{m}r^{2}$$

$$=\sum_{r=1}^{m}\left \{ 8r^{3}-12r^{2}+6r-1 \right \}+12\sum_{r=1}^{m}r^{2}$$

$$=8\sum_{r=1}^{m}r^{3}-12\sum_{r=1}^{m}r^{2}+6\sum_{r=1}^{m}r-\sum_{r=1}^{m}1+12 \sum_{r=1}^{m}r^{2}$$

$$=8\sum_{r=1}^{m}r^{3}+6\sum_{r=1}^{m}r-\sum_{r=1}^{m}1$$

$$=8.\displaystyle \frac{m^{2}(m+1)^{2}}{4}+6\displaystyle \frac{m(m+1)}{2}-m$$

$$=2m^{2}(m+1)^{2}+3m(m+1)-m$$

$$=m[2m^{3}+4m^{2}+5m+2]$$

$$=\displaystyle \frac{n}{2}\left [ 2\left ( \displaystyle \frac{n}{2} \right )^{3}+4\left ( \displaystyle \frac{n}{2} \right )^{2}+5\left ( \displaystyle \frac{n}{2}\right )+2 \right ]$$  $$(\because m=\displaystyle \frac {n}{2})$$

$$=\displaystyle \frac{n}{8}(n^{3}+4n^{2}+10n+8)$$

Since, $$S_{ n }$$ denote the sum of cubes of the first $$n$$ natural numbers.
Therefore, $$S_{ n }=\dfrac { { n }^{ 2 }{ \left( n+1 \right)  }^{ 2 } }{ 4 } $$
Since, $$s_{ n }$$ denote the sum of the first n natural numbers.
Therefore, $$s_{ n }=\dfrac { { n }{ \left( n+1 \right)  } }{ 2 } $$

Now, $${ T }_{ n }=\sum _{ r=1 }^{ n } \dfrac { S_{ r } }{ s_{ r } } =\sum _{ r=1 }^{ n } \dfrac { r\left( r+1 \right)  }{ 2 } $$

$$\Rightarrow { T }_{ n }=\dfrac { 1 }{ 2 } \left\{ \sum _{ r=1 }^{ n }{ { r }^{ 2 } } +\sum _{ r=1 }^{ n } r \right\} =\dfrac { { n }{ \left( n+1 \right)  }{ \left( 2n+1 \right)  } }{ 12 } +\dfrac { { n }{ \left( n+1 \right)  } }{ 4 } $$

$$\Rightarrow { T }_{ n }=\dfrac { n\left( n+1 \right) \left( n+2 \right)  }{ 6 } $$

$${ a }_{ n+1 }={ a }_{ n }+4n+3$$

$$V_r = \dfrac{r}{2}(r+(r-1)(2r-1))$$

We find that
$$S=1.2+1.3+1.4+....+2.3+2.4+.......+3.4+3.5+.....+(n-1)n \cdots (1)$$
$$\because (1+2+3+...+(n-1)+n)^{2}=1^{2}+2^{2}+3^{2}+...(n-1)^{2}+n^{2}$$
$$+2(1.2+1.3+1.4+....+2.3+2.4+.......+3.4+3.5+.....+(n-1)n )$$    from $$(1)$$
$$(\sum n)^{2}=\sum n^{2}+2S$$
$$\Rightarrow S=\displaystyle \frac{(\sum n)^{2}-\sum n^{2}}{2}$$

$$=\displaystyle \frac{\left \{ \displaystyle \frac{n(n+1)}{2} \right \}^{2}-\displaystyle \frac{n(n+1)(2n+1)}{6}}{2}$$
$$=\displaystyle \frac { \displaystyle \frac{n^{2}(n+1)}{4}^{2}-\displaystyle \frac{n(n+1)(2n+1)}{6}}{2}$$
$$=\displaystyle \frac{n(n+1)}{24}\left \{ 3n(n+1)+2(2n+1) \right \}$$
$$=\displaystyle \frac{n(n+1)(3n^{2}-n-2)}{24}$$
$$=\displaystyle \frac{(n+1)n(n-1)(3n+2)}{24}$$

Let $$S=\sum _{ r=1 }^{ n } \left\{ \left( 2r-1 \right) a+\dfrac { 1 }{ b^{ r } }  \right\} $$

$$\Rightarrow S=2a\sum _{ r=1 }^{ n } r-a\sum _{ r=1 }^{ n } +\sum _{ r=1 }^{ n } \left( \dfrac { 1 }{ b^{ r } }  \right) $$

$$\Rightarrow S=an\left( n+1 \right) -an+\dfrac { 1 }{ b } \left[ \dfrac { { \left( 1/b \right)  }^{ n }-1 }{ \left( 1/b \right) -1 }  \right] $$

$$\Rightarrow S=a{ n }^{ 2 }+\dfrac { { b }^{ n }-1 }{ { b }^{ n }\left( b-1 \right)  } $$

Ans: B

Let $$S=11+23+45+87+...$$up to $$n$$ terms
$$\Rightarrow \ S=\left( 5.2+1 \right) +\left( 5.{ 2 }^{ 2 }+3 \right) +\left( 5.{ 2 }^{ 3 }+5 \right) +\left( 5.{ 2 }^{ 4 }+7 \right) +...$$up to $$n$$ terms

Now, $${ a }_{ n }=5{ .2 }^{ n }+2n-1$$
Therefore, $$S=\sum _{ n=1 }^{ n }{ { a }_{ n } } =\sum _{ n=1 }^{ n }{ \left( 5{ .2 }^{ n }+2n-1 \right)  } $$
$$\Rightarrow S=5\sum _{ n=1 }^{ n }{ { 2 }^{ n } } +2\sum _{ n=1 }^{ n }{ n } -\sum _{ n=1 }^{ n }{ 1 } =\dfrac { 5.2\left( { 2 }^{ n }-1 \right)  }{ 2-1 } +\dfrac { 2n\left( n+1 \right)  }{ 2 } -n$$
$$\Rightarrow S=10\left( { 2 }^{ n }-1 \right) +{ n }^{ 2 }$$

nth term $$T_n = 1^2+2^2+3^2+.....+n^2=\sum n^2=\cfrac{1}{6}n(n+1)(2n+1)=\cfrac{1}{6}(2n^3+3n^2+n)$$

Let, $$S=1^{ 3 }-2^{ 3 }+3^{ 3 }-4^{ 3 }+...+9^{ 3 }$$

$$\Rightarrow S=\left( 1^{ 3 }+3^{ 3 }+5^{ 3 }+...+9^{ 3 } \right) -\left( 2^{ 3 }+4^{ 3 }+6^{ 3 }+8^{ 3 } \right) $$

$$\Rightarrow S=\left( 1^{ 3 }+3^{ 3 }+5^{ 3 }+...+9^{ 3 } \right) -2^{ 3 }\left( 1^{ 3 }+2^{ 3 }+3^{ 3 }+4^{ 3 } \right) $$

$$\Rightarrow S=\sum _{ n=1 }^{ 5 }{ { \left( 2n-1 \right)  }^{ 3 } } -2^{ 3 }\sum _{ n=1 }^{ 4 }{ { n }^{ 3 } } $$

$$\Rightarrow S=8\sum _{ n=1 }^{ 5 }{ { n }^{ 3 } } -12\sum _{ n=1 }^{ 5 }{ { n }^{ 2 } } +6\sum _{ n=1 }^{ 5 }{ { n } } -\sum _{ n=1 }^{ 5 }{ 1 } -2^{ 3 }\sum _{ n=1 }^{ 4 }{ { n }^{ 3 } } $$

Consider
$$\displaystyle \left ( b_{1}+b_{2}+b_{3}+...+b_{n} \right )^{2}=b_{1}^{2}+b_{2}^{2}+...+b_{n}^{2}+2\sum_{i= j}b_ib_j $$
taking $$\displaystyle b_{1}=1,b_{2}=2,...,b_{n}=n$$
$$\displaystyle \therefore \left ( 1+2+3+...+n \right )^{2}=1^{2}+2^{2}+3^{2}+...+n^{2}+2\displaystyle \sum $$(Product of number taken two at a time)
$$\Rightarrow 2 \sum{b_ib_j} \displaystyle =\left ( 1+2+3+...+n \right )^{2}-\sum_{n=1}^{n}n^{2}\displaystyle=\frac{n^{2}\left ( n+1 \right )^{2}}{4}-\frac{n\left ( n+1 \right
)\left ( 2n+1 \right )}{6}=\frac{n\left ( n^{2}-1 \right )\left ( 3n+2
\right )}{12}$$
$$\Rightarrow \sum{b_ib_j} =\dfrac{n\left ( n^{2}-1 \right )\left ( 3n+2
\right )}{24} $$

$$S=2.3.1+3.4.4+4.5.7\\ S=\sum _{ r=1 }^{ n }{ \left( r+1 \right) \left( r+2 \right) \left( 3r-2 \right)  } =\sum _{ r=1 }^{ n }{ \left( r+1 \right) \left( { 3r }^{ 2 }+4r-4 \right)  } $$
$$\displaystyle =\sum _{ r=1 }^{ n }{ \left( { 3r }^{ 3 }+{ 7r }^{ 2 }-4 \right)  } =3{ \left( \frac { n\left( n+1 \right)  }{ 2 }  \right)  }^{ 2 }+7\left( \frac { n\left( n+1 \right) \left( 2n+1 \right)  }{ 6 }  \right) -4$$
$$\displaystyle =\frac { n }{ 12 } \left[ { 9n }^{ 3 }+{ 46n }^{ 2 }+51n-34 \right] $$

Let $$a_1=c$$. Then $$a_2=a_1+1=c+1$$,$$a_4=a_2+2=c+3$$. Since the sequence is increasing,it follows that $$a_3=c+2$$. We Prove that $$a_n=c+n1$$ for all $$ n1$$. Indeed, if $$n=2k$$ for some integer $$k$$ this follows by induction on $$k$$. Suppose that $$a_{2^k}=c+2^k1$$. Then $$a_{2^{k+1}}=a_{2.2^k}=a_{2^k}+2^k=c+2^{k+1}1$$. If $$2^k<n<2^{k+1}$$, then $$c+2^{k}1=a_{2^k}<a_{2^k+1}<...<a_n<...<a_{2^k+1}=c+2^{k+1}1$$ add this is possible only if $$a_n=c+n1$$.

Next we prove that $$c=1$$. Suppose that $$c2$$ and let $$p<q$$ be two consecutive prime numbers greater than $$c$$. We have $$a_{qc+1}=c+qc=q$$, hence $$qc+1$$ is a prime and clearly $$qc+1p$$. It follows that for any consecutive prime numbers $$p<q$$ we have $$qc+1$$. The numbers $$(c+1)!+2,(c+1)!+3...,(c+1)!+c+1$$ are all composite, hence if $$p$$ and $$q$$ are the consecutive primes such that $$p<(c+1)!+2<(c+1)!+c+1<q$$ then $$qp>c1$$, a contradiction. It follows that $$c=0$$ and $$a_n=n$$ for all $$n1$$.

$$\displaystyle \sum_{r=1}^{n}r(n-r)^{3}= \sum_{r=1}^{n}(n-r)r^{3}$$
$$=\displaystyle \sum_{r=1}^{n}\left ( nr^{3}-r^{4} \right )$$
$$=\displaystyle n \sum_{r=1}^{n} r^{3}-\sum_{r=1}^{n} r^{4}=n \left \{ \frac{n(n+1)}{2}  \right \}^{2}-f(n)$$

Ans: A

General term of the given series is,
$$\displaystyle T_n =\frac{\sum_{k=1}^n k^2}{\sum_{k=1}^n k}=\frac{\dfrac{1}{6}n(n+1)(2n+1)}{\dfrac{1}{2}n(n+1)}=\frac{1}{3}(2n+1)$$
The required summation is,
$$\displaystyle \sum_{k=1}^nT_k=\frac{1}{3}\left(2\sum_{k=1}^n k+\sum_{k=1}^n 1 \right)=\frac{1}{3}[n(n+1)+n]=\frac{1}{3}n(n+2)$$
Hence option 'D' is correct choice.

$$S={ 1 }^{ 2 }+\left( { 1 }^{ 2 }+{ 2 }^{ 2 } \right) +\left( { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } \right) +...$$

Let   $$S =1+3+6+10+..............+T_n$$ ........(1)
Also $$S =        1+3 + 6+ 10+..........+T_{n-1} +T_n$$ ........(2)
Now by $$(1)-(2)$$
$$0=1+2+3+.........+n-T_n$$
$$\Rightarrow T_n = 1+2+3......+n=\cfrac{1}{2}n(n+1)$$
Hence required summation is,
$$\displaystyle \sum_{k=1}^nT_k=\frac{1}{2}\left(\sum_{k=1}^nk^2+\sum_{k=1}^n k  \right)$$
$$\displaystyle =\frac{1}{2}\left(\frac{1}{6}n(n+1)(2n+1)+\frac{1}{2}n(n+1) \right)$$
$$\quad =\displaystyle \frac{1}{6}n(n+1)(n+2)$$

Sum $$=\displaystyle \sum_{r=1}^{n}r\left ( r+1 \right )\left ( 2r+3 \right )=\sum_{r=1}^{n}(2r^3+5r^2+3r)$$

$$T_n = 1 + a + a^{2} +  .....  + a^{n - 1} = \displaystyle  \frac{1 - a^{n}}{1 - a}$$

$$S_n = \Sigma T_n =\displaystyle \frac{1}{(1 -a)} [\Sigma 1 - \Sigma a^{n}]$$

$$= \displaystyle \frac{1}{(1 - a)} [n - (a + a^{2} + a^{3} + ..... a^{n})]$$

$$= \displaystyle \frac{1}{(1 - a)} \left[n - \frac{a(1 - a^{n})}{(1 - a)} \right] = \frac{n}{1 - a} - \frac{a(1 - a^{n})}{(1 - a)^{2}}$$

$${37\times3=37\times(3\times1)=111}$$
$${37\times 6=37\times(3\times2)=222}$$
$${37\times9=37\times(3\times3)=333}$$


$${37\times27=37\times(3\times9)=999}$$

We have,

 Given expression can be expanded as $$\displaystyle \left ( 1-\frac{1}{4} \right )\left ( 1-\frac{1}{9} \right )\left ( 1-\frac{1}{16} \right )\left ( 1-\frac{1}{25} \right )$$............$$\displaystyle \left ( 1-\frac{1}{81} \right )\left ( 1-\frac{1}{100} \right )$$
$$\displaystyle =\frac{3^{1}}{4_{1}}\times \frac{8^{2^{1}}}{9_{8_{1}}}\times \frac{15^{5^{1}}}{25_{5_{1}}}\times \frac{24^{3^{1}}}{36_{12_{1}}}\times \frac{48^{4^{1}}}{49_{7_{1}}}$$
$$\displaystyle =\frac{63^{7^{1}}}{64_{16^{1}}}\times \frac{80^{5^{1}}}{81_{9_{1}}}\times \frac{99^{11}}{100_{20}}=\frac{11}{20}$$

$$\displaystyle \sum_{s=1}^{n}\, \left \{ \displaystyle \sum_{r=1}^{s}r \right \}\, =\, \displaystyle \sum_{s=1}^{n}\, \displaystyle \frac {s(s+1)}{2}$$

$$=\, \displaystyle \frac {1}{2}\, \left [ \displaystyle \frac {n(n+1)(2n+1)}{6}\, +\, \displaystyle \frac {n(n+1)}{2} \right ]$$

$$=\, \displaystyle \frac {n(n+1)}{4}\, \left [ \displaystyle \frac {2n+1+3}{3} \right ]\, =\, \displaystyle \frac {1}{12}\, n(n+1)(2n+4)$$

$$\Rightarrow\, a\, =\, \displaystyle \frac {1}{6},\, b\, =\, \displaystyle \frac {1}{2},\, c\, =\, \displaystyle \frac {1}{3}$$

$$\Rightarrow a+b+c=1$$

Let $$S_{10}=4 + 8 + 16 + 32 + ........ $$ upto $$10 $$ terms
             $$=4(1+2+4+8+.........$$.upto $$10$$ terms
Clearly, $$1+2+2^2+2^3+...$$ is an GP with first term $$=1$$ and common ratio $$=2$$.
$$\therefore S_{10}=4[\displaystyle \frac{1(2^{10}-1)}{2-1}]=4(2^{10}-1)$$

$$\displaystyle\sum _{ q=1 }^{ 10 }{ \left( \sin { \frac { 2q\pi  }{ 11 } -i } \sin { \frac { 2q\pi  }{ 11 }  }  \right)  } $$

Formula used: Sum of infinite GP
$$S_\infty=a+ar+ar^2+ar^3+...=\dfrac{a}{1-r}$$, where $$|r|<1$$
Given, $$1\, +\, \displaystyle {\frac{1}{3}\, +\, \frac{1}{9}\, +\, \frac{1}{27}\, +\, .......}$$
Clearly, it is a GP with first term $$a=1$$ and common difference $$d=\frac13$$
$$\therefore1\, +\, \displaystyle {\frac{1}{3}\, +\, \frac{1}{9}\, +\, \frac{1}{27}\, +\, .......}=\frac{1}{1-\frac13}=\frac32$$
Hence, option B is correct.

$$\displaystyle S=\frac { { 1 }^{ 2 } }{ 1 } +\frac { { 1 }^{ 2 }+{ 2 }^{ 2 } }{ 1+2 } +\frac { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } }{ 1+2+3 } +...$$

Clearly required ratio is $$=\cfrac{\sum n^3}{\sum n^2}=\cfrac{\cfrac{n^2(n+1)^2}{4}}{\cfrac{  n(n+1) (2n+1)}{6}}=\cfrac{3n(n+1)}{2(2n+1)}$$
Substituting $$n =12$$ we get required value $$=\cfrac{234}{25}$$
Hence, option 'A' is correct.