Sequences and Series 2 Question 10
11.
If $S _n=\sum _2^{4 n}(-1)^{\frac{k(k+1)}{2}} k^{2}$. Then, $S _n$ can take value(s)
(a) 1056
(b) 1088
(c) 1120
(d) 1332
(2013 Adv.)
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Answer:
Correct Answer: 11. (a,d)
Solution:
- PLAN: Convert it into differences and use sum of nterms of an AP,
$ \text { i.e. } \quad S _n=\frac{n}{2}[2 a+(n-1) d] $
Now, $S _n=\sum _{k=1}^{4 n}(-1)^{\frac{k(k+1)}{2}} \cdot k^{2}$
$ \begin{aligned} & =-(1)^{2}-2^{2}+3^{2}+4^{2}-5^{2}-6^{2}+7^{2}+8^{2}+\ldots \\ & =\left(3^{2}-1^{2}\right)+\left(4^{2}-2^{2}\right)+\left(7^{2}-5^{2}\right)+\left(8^{2}-6^{2}\right)+\ldots \\ & =\underbrace{2{(4+6+12+\ldots)} _{n \text { terms }}+\underbrace{(6+14+22+\ldots)}} _{n \text { terms }} \\ & =2 \frac{n}{2}{2 \times 4+(n-1) 8}+\frac{n}{2}{2 \times 6+(n-1) 8} \\ & =2[n(4+4 n-4)+n(6+4 n-4)] \\ & =2\left[4 n^{2}+4 n^{2}+2 n\right]=4 n(4 n+1) \end{aligned} $
Here, $1056=32 \times 33,1088=32 \times 34$,
$ 1120=32 \times 35,1332=36 \times 37 $
1056 and 1332 are possible answers.
