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

The sum of the series $$1^2 - 2^2 + 3^2 - 4^2 + .......... - 1000^2 + 1001^2$$ is
  • $$-501\times1001$$
  • $$501\times1001$$
  • $$499\times1001$$
  • $$-499\times1001$$
$$\displaystyle \frac{1.2^{2}+2.3^{2}+3.4^{2}+.\cdot.\cdot.\cdot.\cdot+n(n+1)^{2}}{1^{2}.2+2^{2}.3+3^{2}.4++n^{2}(n+1)}=$$
  • $$\displaystyle \frac{3n+1}{3n+5}$$
  • $$\displaystyle \frac{3n+5}{3n+1}$$
  • $$(3n+1)(3n+5)$$
  • $$\displaystyle \frac{3n+5}{3n+7}$$
 $$1^{2}+1+2^{2}+2+3^{2}+3+\ldots..+n^{2}+n=$$
  • $$\displaystyle \frac{(n+1)(n+2)}{3}$$
  • $$\displaystyle \frac{n(n+1)(n+2)}{3}$$
  • $$\displaystyle \frac{n(n+2)(n+3)}{6}$$
  • $$\displaystyle \frac{(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{4}$$
The sum of $$7$$ terms of the series $$1^2-2^2+3^2-4^2+5^2-6^2+...$$ is 
  • $$-21$$
  • $$15$$
  • $$28$$
  • $$-35$$
$$1^3+1^2+1+2^{3}+2^{2}+2+3^{3}+3^{2}+3+ ...  3\mathrm{n} $$ terms $$=$$ 
  • $$\displaystyle \frac{n(n+1)(n^{2}+12n+5)}{12}$$
  • $$\displaystyle \frac{n(n+1)(3n^{2}+7n+8)}{12}$$
  • $$\displaystyle \frac{n(n+1)(n+2)(n^{2}+5n+6)}{12}$$
  • $$\displaystyle \frac{(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{4}$$
The sum $$\mathrm{V}_{1}+\mathrm{V}_{2}+\ldots+\mathrm{V}_{\mathrm{n}}$$ is 

  • $$\displaystyle \frac{1}{12}\mathrm{n}(\mathrm{n}+1)(3\mathrm{n}^{2}-\mathrm{n}+1)$$
  • $$\displaystyle \frac{1}{12}\mathrm{n}(\mathrm{n}+1)(3\mathrm{n}^{2}+\mathrm{n}+2)$$
  • $$\displaystyle \frac{1}{2}\mathrm{n}(2\mathrm{n}^{2}-\mathrm{n}+1)$$
  • $$\displaystyle \frac{1}{3}(2\mathrm{n}^{3}-2\mathrm{n}+3)$$
The sum of the infinite series, $$\dfrac{1}{1.3}+\dfrac{2}{3.7}+\dfrac{3}{7.13}+\dfrac{4}{13.21}+ ..........$$ is

  • $$\dfrac{1}{2}$$
  • $$\dfrac{25}{24}$$
  • $$\dfrac{25}{54}$$
  • $$\dfrac{125}{252}$$
Find $$\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n-1)!}$$
  • $$\displaystyle \frac{1}{2}(e+e^{-1})$$
  • $$\displaystyle \frac{1}{2}(e-e^{-1})$$
  • $$e+e^{-1}$$
  • $$\mathrm{e}-\mathrm{e}^{-1}$$
$$\displaystyle \frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3}+\ldots+n$$ terms $$=$$
  • $$\displaystyle \frac{n(n+3)}{4}$$
  • $$\displaystyle \frac{n(n+3)}{5}$$
  • $$\displaystyle \frac{n(n+2)}{3}$$
  • $$\displaystyle \frac{n(n+5)}{6}$$
Evaluate $$1.6+2.9+3.12+...+\mathrm{n}(3\mathrm{n}+3)=$$
  • $$\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)$$
  • $$(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)$$
  • $$(\mathrm{n}+2)(\mathrm{n}+3)(\mathrm{n}+4)$$
  • $$(n-1) n(n+1)$$
$$2.4+4.7+6.10+...(n-1)$$ terms
  • $$2\mathrm{n}^{3}-2\mathrm{n}^{2}$$
  • $$\displaystyle \frac{n^{3}+3n^{2}+1}{6}$$
  • $$2n^{3}+2n$$
  • $$2\mathrm{n}^{3}-\mathrm{n}^{2}$$
The sum of the first $$n$$ terms of the series $$1^2+2.2^2+3^2+2.4^2+5^2+2.6^2+.....$$ is $$\dfrac {n(n+1)^2}{2}$$, when $$n$$ is even,when $$n$$ is odd, the sum is
  • $$\dfrac {n^2(n+1)}{2}$$
  • $$\left[ \dfrac {n(n+1)}{2}\right]^{2}$$
  • $$\dfrac {3n(n+1)}{2}$$
  • $$\dfrac {n(n+1)^{2}}{4}$$
$$x = (-1)^{a^1} + (-1)^{a^2} + .....(-1)^{a^{1006}}$$
$$ y = (-1)^{a^{1007}} + (-1)^{a^{1008}} + ...+(-1)^{a^{2013}}$$
Then which of the following is true?
  • $$(-1)^{x} = 1; (-1)^{y} =1$$
  • $$(-1)^{x} = 1; (-1)^{y} = -1$$
  • $$ (-1)^{x} = -1; (-1)^{y} =1$$
  • $$(-1)^{x} = -1; (-1)^{y} = -1$$
What number corresponds to the fifth missing planet ?
  • 28
  • 78
  • 32
  • 36
State true or false:
Given sequence $$15, 30, 60, 120, 240$$ is in G.P.
  • True
  • False
Let a sequence $$\left \{a_n\right \}$$ be defined by $$a_n=\dfrac {1}{n+1}+\dfrac {1}{n+2}+\dfrac {1}{n+3}+.....+\dfrac {1}{3n}$$, then
  • $$a_2=\dfrac {7}{12}$$
  • $$a_2=\dfrac {19}{20}$$
  • $$a_{n+1}-a_n=\dfrac {9n+5}{(3n+1)(3n+2)(3n+3)}$$
  • $$a_{n+1}-a_n=\dfrac {-2}{3(n+1)}$$
In the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, .......... starting with the third terms, each term is the sum of the two preceding terms : $$2=1+1, 3=1+2, 5=2+3$$, and so on.
Choose the correct options -
  • Eighth term in Fibonacci sequence is 21
  • Tenth term in Fibonacci sequence is 55
  • Twelfth term in Fibonacci sequence is 144
  • Twelfth term in Fibonacci sequence is 112
State true or false:
A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.
Parts of red pigment
1
4
7
12
20
Parts of base
8
...
...
...
...

Answer: 
Parts of red pigment
1
4
7
12
20
Parts of base
8
325696150
  • True
  • False
The value of $$\alpha$$ for which $$\sum_{n=1}^{\infty}A_n=\frac {8}{3}$$ is-
114347_20b59845404741bc87d02969783237ad.png
  • 1/3, 2/3
  • 1/4, 3/4
  • 1/5, 4/5
  • 1/2
Out of nine cells of a square, one cell is left blank and in the rest of the cells, numbers are written follow some rule. Get the rule and find out the proper option for the blank cell 
2
72
56
?
0
42
12
20
30
  • 4
  • 6
  • 8
  • 10
Which of the following is not a G.P.?
  • $$2, 4, 6, 8....$$
  • $$5, 25, 125, 625....$$
  • $$1.5, 3.0, 6.0, 12.0....$$
  • $$8, 16, 24, 32, ....$$
The value of $$\sum_{n=1}^{10}\int_{0}^{n}xdx$$ is
  • an even integer
  • an odd integer
  • a rational number
  • an irrational number
If n is an odd integer greater than or equal to 1 then the value of $$n^{3}-\left ( n-1 \right )^{3}+\left ( n-2 \right )^{3}-...+\left ( -1 \right )^{n-1}.1^{3}$$ is
  • $$\dfrac{\left ( n+1 \right )^{2}.\left ( 2n-1 \right )}{4}$$
  • $$\dfrac{\left ( n-1 \right )^{2}.\left ( 2n-1 \right )}{4}$$
  • $$\dfrac{\left ( n+1 \right )^{2}.\left ( 2n+1 \right )}{4}$$
  • none of these
Find the sum of first $$n$$ terms of the series.
$$1^{3}+3\times2^{2}+3^{3}+3\times4^{2}+5^{3}+3\times6^{2}+....$$ If $$n$$ is even.
  • $$\displaystyle \frac{n}{8}(n^{3}+4n^{2}+10n+6)$$
  • $$\displaystyle \frac{n}{8}(n^{2}+4n^{2}+10n+6)$$
  • $$\displaystyle \frac{n}{8}(n^{3}+4n^{2}+10n+8)$$
  • $$\displaystyle \frac{n}{8}(n^{2}+4n^{2}+10n+8)$$
Let $$S_{n}$$ denote the sum of cubes of the first $$n$$ natural numbers and $$s_{n}$$ denote the sum of the first $$n$$ natural numbers. Then $$\sum_{r=1}^{n}\dfrac{S_{r}}{S_{r}}$$ is equal to 
  • $$\dfrac{n\left ( n+1 \right )\left ( n+2 \right )}{6}$$
  • $$\dfrac{n\left ( n+1 \right )}{2}$$
  • $$\dfrac{n^{2}+3n+2}{6}$$
  • none of these
Let a sequence be defined by $$a_{1}= 0$$ and $$a_{n+1}= a_{n}+4n+3$$ for all $$n\geq 1 (n\epsilon N)$$

The value of $$a_{k}$$ in terms of k is $$(k\in N)$$

  • $$(k-1)(2k+3)$$
  • $$(k-1)(3k+1)$$
  • $$(k-1)(k+3)$$
  • $$(k-1)(2k+5)$$
Let $$\displaystyle V_{r}$$ denote the sum of the first r terms of an arithmetic progression (AP) whose first term is r and the common difference is $$\displaystyle (2r-1).$$ Let  $$\displaystyle T_{r}=V_{r+1}-V_{r}-2$$ and  $$\displaystyle Q_{r}=T_{r+1}-T_{r}$$ for $$r=1, 2, ...$$
The sum  $$\displaystyle V_{1}+V_{2}+... +V_{n}$$ is
  • $$\displaystyle \frac{1}{12}n\left ( n+1 \right )\left ( 3n^{2}-n+1 \right )$$
  • $$\displaystyle \frac{1}{12}n\left ( n+1 \right )\left ( 3n^{2}+n+2 \right )$$
  • $$\displaystyle \frac{1}{2}n\left ( 2n^{2}-n+1 \right )$$
  • $$\displaystyle \frac{1}{3}\left ( 2n^{3}-2n+3 \right )$$
Find the sum of the products of every pair of the first $$n$$ natural numbers.
  • $$S=\displaystyle \frac{(n-1)^2(3n+2)}{24}$$
  • $$S=\displaystyle \frac{n(n+1)(n-1)(3n-2)}{24}$$
  • $$S=\displaystyle \frac{n(n-1)^2(3n+2)}{24}$$
  • $$S=\displaystyle \frac{n(n+1)(n-1)(3n+2)}{24}$$
The value of $$\sum_{r=1}^{n}\left \{ \left ( 2r-1 \right )a+\dfrac{1}{b^{r}} \right \}$$ is equal to 
  • $$an^{2}+\dfrac{b^{n-1}-1}{b^{n-1}\left ( b-1 \right )}$$
  • $$an^{2}+\dfrac{b^{n}-1}{b^{n}\left ( b-1 \right )}$$
  • $$an^{3}+\dfrac{b^{n-1}-1}{b^{n}\left ( b-1 \right )}$$
  • none of these
Sum of the series $$11+23+45+87 ...$$ up to $$n$$ terms
  • $$10(2^n-1)+n^2$$
  • $$10(2^n+1)+n^2$$
  • $$20(2^n-n)+n^2$$
  • None of these
$$\displaystyle 1^{2}+\left ( 1^{2}+2^{2} \right )+\left (1^{2}+2^{2}+3^{2} \right )+...$$ to $$n$$ terms.
  • $$\dfrac{1}{12}n(n+1)(2n^3+3n^2+n)$$
  • $$\dfrac{1}{12}n(n+1)(3n^3+2n^2+n)$$
  • $$\dfrac{1}{6}n(n+1)(2n+1)$$
  • $$\dfrac{1}{4}n^2(n+1)^2$$
$$\displaystyle 1^{3}-2^{3}+3^{3}-4^{3}+...+9^{3}$$ equals
  • 425
  • 400
  • 405
  • 395
The sum of all possible product of $$1^{st}$$ $$n$$ natural numbers taken two at a time is
  • $$\displaystyle \frac{n\left ( n+1 \right )\left ( n+2 \right )\left ( 2n+3 \right )}{24}$$
  • $$\displaystyle \frac{n\left ( n^{2}-1 \right )\left ( 3n+2 \right )}{24}$$
  • $$\displaystyle \frac{n\left ( n^{2}+1 \right )\left ( 3n+2 \right )}{6}$$
  • $$\displaystyle \frac{n\left ( n^{2}-1 \right )\left ( 3n+2 \right )}{6}$$
Sum of the series $$2.3.1+3.4.4+4.5.7+...$$ up to $$n$$ terms is
  • $$\displaystyle \frac{n}{12}\left [ 9n^{3}+46n^{2}+51n-34 \right ]$$
  • $$\displaystyle \frac{n}{6}\left [ 9n^{3}+46n^{2}+51n-34 \right ]$$
  • $$\displaystyle \frac{\left ( n+1 \right )}{6}\left [ 9n^{3}+46n^{2}-34 \right ]$$
  • $$\displaystyle \frac{n}{12}\left [ 9n^{3}+46n^{2}-34 \right ]$$
Let $$\displaystyle \left ( a_{n} \right )n\geq 1$$ be an increasing sequence of positive integers such that 1.$$\displaystyle a_{2n}=a_{n}+n$$ for all $$\displaystyle n\geq 1$$ 2.if $$\displaystyle a_{n}$$ is prime, then n is a prime. Prove that $$\displaystyle a_{n}=n,$$ for all $$\displaystyle n\geq 1.$$
  • $$\displaystyle n\geq 1.$$
  • $$\displaystyle n\ne 1.$$
  • $$\displaystyle n= 1.$$
  • None of the above
  • Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
  • Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
  • Assertion is correct but Reason is incorrect
  • Both Assertion and Reason are incorrect
$$\displaystyle \frac{1^{2}}{1} + \frac{1^{2} + 2^{2}}{1 + 2} + \frac{1^{2} + 2^{2} + 3^{2}}{1 + 2 + 3} + .....$$ upto n terms is
  • $$\displaystyle \frac{1}P{3} (2n + 1)$$
  • $$\displaystyle \frac{1}{3} n^{2}$$
  • $$\displaystyle \frac{1}{3} (n + 2)$$
  • $$\displaystyle \frac{1}{3} n(n + 2)$$
The sum of n terms of $$1^{2}\, +\, (1^{2}\, +\, 2^{2})\, +\, (1^{2}\, +\, 2^{2}\, +\, 3^{2})\, +\, ....$$.
  • $$\displaystyle \frac{n(n\, +\,a)\, (2n\, +\, 1)}{6}$$
  • $$\displaystyle \frac{n(n\, +\, 1)\, (2n\, -\, 1)}{6}$$
  • $$\displaystyle \frac{1}{12}\, n(n\, +\, 1)^{2}\, (n\, +\, 2)$$
  • $$\displaystyle \frac{1}{12}n^{2}(n\, +\, 1)^{2}$$
1 + 3 + 6 + 10 + .... upto n terms is equal to
  • $$\displaystyle \frac{1}{3} n(n + 1)(n + 2)$$
  • $$\displaystyle \frac{1}{6} n(n + 1)(n + 2)$$
  • $$\displaystyle \frac{1}{12} n(n + 2)(n + 3)$$
  • $$\displaystyle \frac{1}{12} n(n + 1)(n + 2)$$
Determine the fourth degree expression in n which is equal to $$\displaystyle \sum_{r=1}^{n}r\left ( r+1 \right )\left ( 2r+3 \right ).$$
  • $$\displaystyle \frac{1}{6}\left ( 3n^{4}+16n^{3}+27n^{2}+14n \right ).$$
  • $$\displaystyle \frac{1}{6}\left ( 3n^{4}+9n^{3}+27n^{2}+36n \right ).$$
  • $$\displaystyle \frac{1}{6}\left ( 3n^{4}+36n^{3}+27n^{2}+9n \right ).$$
  • $$\displaystyle \frac{1}{6}\left ( 3n^{4}+14n^{3}+27n^{2}+16n \right ).$$
The sum of n terms of the series
$$1 + (1 + a) + (1 + a + a^{2}) + (1 + a + a^{2} + a^{3}) + .....$$, is
  • $$\displaystyle \frac{n}{1 - a} - \frac{a(1 - a^{n})}{(1 - a)^{2}}$$
  • $$\displaystyle \frac{n}{1 - a} + \frac{a(1 - a^{n})}{(1 - a)^{2}}$$
  • $$\displaystyle \frac{n}{1 - a} - \frac{a(1 + a^{n})}{(1 - a)^{2}}$$
  • $$\displaystyle - \frac {n}{1 - a} + \frac{a(1 - a^{n})}{(1 - a)^{2}}$$
Observe the given multiples of 37.
$${37\times3=111}$$
$${37\times 6 =222}$$
$${37\times9=333}$$
$${37\times12=444}$$-------------------------------
Find the product of $${37\times27}$$
  • $$999$$
  • Greatest  $$3$$ digit number
  • Either A or B
  • Smallest  $$3$$ digit number
Sum to $$20$$ terms of the series $$1.{ 3 }^{ 2 }+2.{ 5 }^{ 2 }+3.{ 7 }^{ 2 }+...$$ is
  • $$178090$$
  • $$168090$$
  • $$188090$$
  • None of these
Find the value of 
$$\displaystyle \left ( 1-\frac{1}{2^{2}} \right )\left ( 1-\frac{1}{3^{3}} \right )\left (1 -\frac{1}{4^{2}} \right )\left ( 1-\frac{1}{5^{2}} \right )........\left ( 1-\frac{1}{9^{2}} \right )\left ( 1-\frac{1}{10^{2}} \right )$$
  • $$\displaystyle \frac{5}{12}$$
  • $$\displaystyle \frac{1}{2}$$
  • $$\displaystyle \frac{11}{20}$$
  • $$\displaystyle \frac{7}{10}$$
If $$\displaystyle \sum_{s=1}^{n}\, \left \{ \displaystyle \sum_{r=1}^{s}r \right \}\, =\, an^3\, +\, bn^2\, +\, cn$$, then find the value of a + b + c.
  • 1
  • 0
  • 2
  • 3
The sum of the series $$4 + 8 + 16 + 32 + .......$$. till $$10$$ terms is
  • $$6\, \times\, 2^{10}\, -\, 1$$
  • $$4\, \times\, 2^{10}\, +\, 1$$
  • $$4(2^{10}\, -\, 1)$$
  • $$4(2^{10}\, +\, 1)$$
$$\displaystyle \sum _{ p=1 }^{ 32 }{ \left( 3p+2 \right) { \left[ \sum _{ q=1 }^{ 10 }{ \left( \sin { \frac { 2q\pi  }{ 11 }  } -i\cos { \frac { 2q\pi  }{ 11 }  }  \right)  }  \right]  }^{ p } } =$$
  • $$8\left( 1-i \right) $$
  • $$16\left( 1-i \right) $$
  • $$48\left( 1-i \right) $$
  • None of these
Find the sum the infinite G.P.:
$$1\, +\, \displaystyle {\frac{1}{3}\, +\, \frac{1}{9}\, +\, \frac{1}{27}\, +\, .......}$$
  • $$\displaystyle \frac{3}{5}$$
  • $$\displaystyle \frac{3}{2}$$
  • $$\displaystyle \frac{49}{27}$$
  • $$\displaystyle \frac{8}{5}$$
The sum of series $$\displaystyle \frac {1^2}{1}+\frac {1^2+2^2}{1+2}+\frac {1^2+2^2+3^2}{1+2+3}+....$$ upto n terms is
  • $$\displaystyle\frac {1}{3}(2n+1)$$
  • $$\displaystyle\frac {1}{3}n^2$$
  • $$\displaystyle\frac {1}{3}(n+2)$$
  • $$\displaystyle\frac {1}{3}n(n+2)$$
$$\dfrac {1^3+2^3+3^3+4^3+........12^3}{1^2+2^2+3^2+4^2+......12^2}=$$
  • $$\dfrac {234}{25}$$
  • $$\dfrac {243}{35}$$
  • $$\dfrac {263}{27}$$
  • None of these
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