Sum of Special Series

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JEE Main 2016

LEVELJEE Main

If the sum of the first ten terms of the series $(1 \frac{3}{5})^{2} + (2 \frac{2}{5})^{2} + (3 \frac{1}{5})^{2} + 4^{2} + (4 \frac{4}{5})^{2} + \dots$ is $\frac{16}{5} m$ , then $m$ is equal to:

The sum of first $9$ terms of the series $\frac{1 ^{3}}{1} + \frac{1 ^{3} + 2 ^{3}}{1 + 3} + \frac{1 ^{3} + 2 ^{3} + 3 ^{3}}{1 + 3 + 5} + \dots$ is

Statement-1: The sum of the series $1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \dots + (361 + 380 + 400)$ is $8000$ . Statement-2: $\sum_{k = 1}^{n} (k^{3} - (k - 1)^{3}) = n^{3}$ , for any natural number $n$ .

(A)

Statement-1 is false, Statement-2 is true.

(B)

Statement-1 is true, Statement-2 is true; statement-2 is a correct explanation for Statement-1.

(C)

Statement-1 is true, Statement-2 is true; statement-2 is not a correct explanation for Statement-1.

(D)

Statement-1 is true, Statement-2 is false.

JEE Main 2008

LEVELBoard

Statement-1 : For every natural number $n \geq 2, \frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n} > n$ . Statement-2 : For every natural number $n \geq 2, n (n + 1) < n + 1$ .

(A)

Statement -1 is false, Statement-2 is true

(B)

Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1

(C)

Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1

(D)

Statement -1 is true, Statement-2 is false

JEE Main 2007

LEVELBoard

The sum of series $\frac{1}{2 !} - \frac{1}{3 !} + \frac{1}{4 !} - \dots$ upto infinity is

(A)

e^{- 2}

(B)

e^{- 1/2}

(C)

e^{- 2}

(D)

e^{- 1}

JEE Main 2005

LEVELJEE Main

The sum of the series $1 + \frac{1}{4.2 !} + \frac{1}{16.4 !} + \frac{1}{64.6 !} + \dots$ ad inf. is

(A)

\frac{e - 1}{e}

(B)

\frac{e + 1}{e}

(C)

\frac{e - 1}{2 e}

(D)

\frac{e + 1}{2 e}

JEE Main 2004

LEVELJEE Main

Let $S (K) = 1 + 3 + 5 + ... + (2 K - 1) = 3 + K^{2}$ . Then which of the following is true

(A)

Principle of mathematical induction can be used to prove the formula

(B)

S (K) \Rightarrow S (K + 1)

(C)

S (K) \neq \Rightarrow S (K + 1)

(D)

S (1)

is correct

JEE Main 2004

LEVELJEE Main

The sum of the first $n$ terms of the series $1^{2} + 2. 2^{2} + 3^{2} + 2. 4^{2} + 5^{2} + 2. 6^{2} + \dots$ is $n (n + 1)^{2} /2$ when $n$ is even. When $n$ is odd the sum is

(A)

[\frac{n ( n + 1 )}{2}]^{2}

(B)

\frac{n ^{2} ( n + 1 )}{2}

(C)

\frac{n ( n + 1 ) ^{2}}{4}

(D)

\frac{3 n ( n + 1 )}{2}

JEE Main 2004

LEVELBoard

The sum of series $\frac{1}{2 !} + \frac{1}{4 !} + \frac{1}{6 !} + \dots$ is

(A)

\frac{e ^{2} - 2}{e}

(B)

\frac{( e - 1 ) ^{2}}{2 e}

(C)

\frac{e ^{2} - 1}{2 e}

(D)

\frac{e ^{2} - 1}{2}

JEE Main 2003

LEVELJEE Main

The sum of the series $\frac{1}{1.2} - \frac{1}{2.3} + \frac{1}{3.4} - \dots$ up to $\infty$ is equal to

(A)

lo g_{e} (\frac{4}{e})

(B)

2 lo g_{e} 2

(C)

lo g_{e} 2 - 1

(D)

lo g_{e} 2

JEE Main 2002

LEVELJEE Main

If $a_{n} = 7 + 7 + 7 + ....$ having $n$ radical signs then by methods of mathematical induction which is true

(A)

a_{n} > 7 \forall n \geq 1

(B)

a_{n} < 7 \forall n \geq 1

(C)

a_{n} < 4 \forall n \geq 1

(D)

a_{n} < 3 \forall n \geq 1

JEE Main 2002

LEVELBoard

$1^{3} - 2^{3} + 3^{3} - 4^{3} + \dots + 9^{3} =$

Let $S_{n} = \sum_{k = 1}^{4 n} (- 1)^{\frac{k ( k + 1 )}{2}} k^{2}$ . Then $S_{n}$ can take value(s)

* Multiple Correct Options

In the sum of first $n$ terms of an A.P. is $c n^{2}$ , then the sum of squares of these $n$ terms is

(A)

\frac{n ( 4 n ^{2} - 1 ) c ^{2}}{6}

(B)

\frac{n ( 4 n ^{2} + 1 ) c ^{2}}{3}

(C)

\frac{n ( 4 n ^{2} - 1 ) c ^{2}}{3}

(D)

\frac{n ( 4 n ^{2} + 1 ) c ^{2}}{6}

JEE Advanced 1999

LEVELJEE Main

For a positive integer $n$ , let $a (n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{( 2 ^{n} ) - 1}$ . Then

* Multiple Correct Options

(A)

a (100) \leq 100

(B)

a (100) > 100

(C)

a (200) \leq 100

(D)

a (200) > 100

JEE Advanced 1988

LEVELJEE Main

Sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is equal to

(A)

2^{n} - n - 1

(B)

1 - 2^{- n}

(C)

n + 2^{- n} - 1

(D)

2^{n} + 1

JEE Advanced 2013

LEVELJEE Main

A pack contains $n$ cards numbered from 1 to $n$ . Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is $k$ , then $k - 20 =$ .........

JEE Advanced 2010

LEVELJEE Main

Let $S_{k}, k = 1, 2, \dots, 100$ , denote the sum of the infinite geometric series whose first term is $\frac{k - 1}{k !}$ and the common ratio is $\frac{1}{k}$ . Then the value of $\frac{10 0 ^{2}}{100 !} + \sum_{k = 1}^{100} ∣ (k^{2} - 3 k + 1) S_{k} ∣$ is \dots.

JEE Advanced 1996

LEVELJEE Main

For any odd integer $n \geq 1$ , $n^{3} - (n - 1)^{3} + \dots + (- 1)^{n - 1} 1^{3} = \dots$ .

JEE Advanced 1988

LEVELJEE Main

The sum of the first $n$ terms of the series $1^{2} + 2. 2^{2} + 3^{2} + 2. 4^{2} + 5^{2} + 2. 6^{2} + \dots$ is $n (n + 1)^{2} /2$ , when $n$ is even. When $n$ is odd, the sum is \dots.

JEE Advanced 1983

LEVELJEE Main

Use mathematical Induction to prove : If $n$ is any odd positive integer, then $n (n^{2} - 1)$ is divisible by 24.

JEE Main 2018 (16 April Shift 1)

LEVELBoard

The sum of the first 20 terms of the series $1 + \frac{3}{2} + \frac{7}{4} + \frac{15}{8} + \frac{31}{16} + \dots$ is :

(A)

39 + \frac{1}{2 ^{19}}

(B)

39 + \frac{1}{2 ^{20}}

(C)

38 + \frac{1}{2 ^{20}}

(D)

38 + \frac{1}{2 ^{19}}

JEE Main 2019 (9 January)

LEVELJEE Main

The sum of the following series $1 + 6 + \frac{9 ( 1 ^{2} + 2 ^{2} + 3 ^{2} )}{7} + \frac{12 ( 1 ^{2} + 2 ^{2} + 3 ^{2} + 4 ^{2} )}{9} + \frac{15 ( 1 ^{2} + 2 ^{2} + ... + 5 ^{2} )}{11} + ...$ up to 15 terms, is:

JEE Main 2019 (12 January)

LEVELBoard

If the sum of the first 15 terms of the series $(3/4)^{3} + (1/2)^{3} + (2/3)^{3} + 3^{3} + (3/4)^{3} + \dots$ is equal to $225 k$ , then $k$ is equal to :

(A)

9

(B)

27

(C)

108

(D)

54

JEE Main 2019 (12 January Shift 1)

LEVELBoard

Let $S_{k} = \frac{1 + 2 + 3 + ... + k}{k}$ . If $S_{1}^{2} + S_{2}^{2} + ... + S_{10}^{2} = \frac{5}{12} A$ , then A is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the sum of the first ten terms of the series (153​)2+(252​)2+(351​)2+42+(454​)2+… is 516​m, then m is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of first 9 terms of the series 113​+1+313+23​+1+3+513+23+33​+… is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Statement-1: The sum of the series 1+(1+2+4)+(4+6+9)+(9+12+16)+⋯+(361+380+400) is 8000. Statement-2: ∑k=1n​(k3−(k−1)3)=n3, for any natural number n.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Statement-1 : For every natural number n≥2,1​1​+2​1​+⋯+n​1​>n​. Statement-2 : For every natural number n≥2,n(n+1)​<n+1.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of series 2!1​−3!1​+4!1​−… upto infinity is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the series 1+4.2!1​+16.4!1​+64.6!1​+… ad inf. is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S(K)=1+3+5+...+(2K−1)=3+K2. Then which of the following is true

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the first n terms of the series 12+2.22+32+2.42+52+2.62+… is n(n+1)2/2 when n is even. When n is odd the sum is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of series 2!1​+4!1​+6!1​+… is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the series 1.21​−2.31​+3.41​−… up to ∞ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If an​=7+7+7+....​​​ having n radical signs then by methods of mathematical induction which is true

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 13−23+33−43+⋯+93=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Sn​=∑k=14n​(−1)2k(k+1)​k2. Then Sn​ can take value(s)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the sum of first n terms of an A.P. is cn2, then the sum of squares of these n terms is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a positive integer n, let a(n)=1+21​+31​+41​+⋯+(2n)−11​. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sum of the first n terms of the series 21​+43​+87​+1615​+… is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For any odd integer n≥1, n3−(n−1)3+⋯+(−1)n−113=….

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the first n terms of the series 12+2.22+32+2.42+52+2.62+… is n(n+1)2/2, when n is even. When n is odd, the sum is \dots.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Use mathematical Induction to prove : If n is any odd positive integer, then n(n2−1) is divisible by 24.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the first 20 terms of the series 1+23​+47​+815​+1631​+… is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the following series 1+6+79(12+22+32)​+912(12+22+32+42)​+1115(12+22+...+52)​+... up to 15 terms, is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the sum of the first 15 terms of the series (3/4)3+(1/2)3+(2/3)3+33+(3/4)3+… is equal to 225k, then k is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Sk​=k1+2+3+...+k​. If S12​+S22​+...+S102​=125​A, then A is equal to :

If the sum of the first ten terms of the series $(1 \frac{3}{5})^{2} + (2 \frac{2}{5})^{2} + (3 \frac{1}{5})^{2} + 4^{2} + (4 \frac{4}{5})^{2} + \dots$ is $\frac{16}{5} m$ , then $m$ is equal to:

The sum of first $9$ terms of the series $\frac{1 ^{3}}{1} + \frac{1 ^{3} + 2 ^{3}}{1 + 3} + \frac{1 ^{3} + 2 ^{3} + 3 ^{3}}{1 + 3 + 5} + \dots$ is

Statement-1: The sum of the series $1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \dots + (361 + 380 + 400)$ is $8000$ . Statement-2: $\sum_{k = 1}^{n} (k^{3} - (k - 1)^{3}) = n^{3}$ , for any natural number $n$ .

Statement-1 : For every natural number $n \geq 2, \frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n} > n$ . Statement-2 : For every natural number $n \geq 2, n (n + 1) < n + 1$ .

The sum of series $\frac{1}{2 !} - \frac{1}{3 !} + \frac{1}{4 !} - \dots$ upto infinity is

The sum of the series $1 + \frac{1}{4.2 !} + \frac{1}{16.4 !} + \frac{1}{64.6 !} + \dots$ ad inf. is

Let $S (K) = 1 + 3 + 5 + ... + (2 K - 1) = 3 + K^{2}$ . Then which of the following is true

The sum of the first $n$ terms of the series $1^{2} + 2. 2^{2} + 3^{2} + 2. 4^{2} + 5^{2} + 2. 6^{2} + \dots$ is $n (n + 1)^{2} /2$ when $n$ is even. When $n$ is odd the sum is

The sum of series $\frac{1}{2 !} + \frac{1}{4 !} + \frac{1}{6 !} + \dots$ is

The sum of the series $\frac{1}{1.2} - \frac{1}{2.3} + \frac{1}{3.4} - \dots$ up to $\infty$ is equal to

If $a_{n} = 7 + 7 + 7 + ....$ having $n$ radical signs then by methods of mathematical induction which is true

$1^{3} - 2^{3} + 3^{3} - 4^{3} + \dots + 9^{3} =$

Let $S_{n} = \sum_{k = 1}^{4 n} (- 1)^{\frac{k ( k + 1 )}{2}} k^{2}$ . Then $S_{n}$ can take value(s)

In the sum of first $n$ terms of an A.P. is $c n^{2}$ , then the sum of squares of these $n$ terms is

For a positive integer $n$ , let $a (n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{( 2 ^{n} ) - 1}$ . Then

Sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is equal to

For any odd integer $n \geq 1$ , $n^{3} - (n - 1)^{3} + \dots + (- 1)^{n - 1} 1^{3} = \dots$ .

The sum of the first $n$ terms of the series $1^{2} + 2. 2^{2} + 3^{2} + 2. 4^{2} + 5^{2} + 2. 6^{2} + \dots$ is $n (n + 1)^{2} /2$ , when $n$ is even. When $n$ is odd, the sum is \dots.

Use mathematical Induction to prove : If $n$ is any odd positive integer, then $n (n^{2} - 1)$ is divisible by 24.

The sum of the first 20 terms of the series $1 + \frac{3}{2} + \frac{7}{4} + \frac{15}{8} + \frac{31}{16} + \dots$ is :

The sum of the following series $1 + 6 + \frac{9 ( 1 ^{2} + 2 ^{2} + 3 ^{2} )}{7} + \frac{12 ( 1 ^{2} + 2 ^{2} + 3 ^{2} + 4 ^{2} )}{9} + \frac{15 ( 1 ^{2} + 2 ^{2} + ... + 5 ^{2} )}{11} + ...$ up to 15 terms, is:

If the sum of the first 15 terms of the series $(3/4)^{3} + (1/2)^{3} + (2/3)^{3} + 3^{3} + (3/4)^{3} + \dots$ is equal to $225 k$ , then $k$ is equal to :

Let $S_{k} = \frac{1 + 2 + 3 + ... + k}{k}$ . If $S_{1}^{2} + S_{2}^{2} + ... + S_{10}^{2} = \frac{5}{12} A$ , then A is equal to :