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Sequences and Series question

2009 · Shift 2 · Q32

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JEE Advanced
Mathematics
Sequences and Series
2009 · Shift 2 · Q32

Sequences and Series question

2009 · Shift 2 · Q32

JEE AdvancedMathematicsSequences and SeriesMCQ+3 / −1

If 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
${{n\left( {4{n^2} - 1} \right){c^2}} \over 6}$
B
${{n\left( {4{n^2} + 1} \right){c^2}} \over 3}$
C
${{n\left( {4{n^2} - 1} \right){c^2}} \over 3}$
D
${{n\left( {4{n^2} + 1} \right){c^2}} \over 6}$

View written solutionFree

Correct answer: C

Let the terms of the A.P. be $a_1,a_2,a_3,\dots$ and let the sum of first $n$ terms be

$S_n=cn^2.$

The $n$ th term of the A.P. is

$a_n=S_n-S_{n-1}.$

Now,

$S_{n-1}=c(n-1)^2=c(n^2-2n+1).$

So,

$a_n=cn^2-c(n-1)^2$ $=c\big(n^2-(n^2-2n+1)\big)$ $=c(2n-1).$

Thus the A.P. is

$c,\ 3c,\ 5c,\dots,(2n-1)c.$

We need the sum of squares of the first $n$ terms:

$\sum_{k=1}^n a_k^2=\sum_{k=1}^n \big(c(2k-1)\big)^2$ $=c^2\sum_{k=1}^n (2k-1)^2.$

Use the standard identity for sum of squares of first $n$ odd numbers:

$\sum_{k=1}^n (2k-1)^2=\frac{n(4n^2-1)}{3}.$

Hence,

$\sum_{k=1}^n a_k^2=c^2\cdot \frac{n(4n^2-1)}{3}.$

Therefore,

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

Comparing with the options:

A: $\dfrac{n(4n^2-1)c^2}{6}$
B: $\dfrac{n(4n^2+1)c^2}{3}$
C: $\dfrac{n(4n^2-1)c^2}{3}$
D: $\dfrac{n(4n^2+1)c^2}{6}$

So the correct option is C.

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