Sequences And Series Ques 45
The sum of the first $n$ terms of the series $1^{2}+2 \cdot 2^{2}+3^{2}+2 \cdot 4^{2}+5^{2}+2 \cdot 6^{2}+\ldots$ is $\frac{n(n+1)^{2}}{2}$, when $n$ is even. When $n$ is odd, the sum is …. .
(1988, 2M)
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Answer:
Correct Answer: 45.[$\frac{n^{2}(n+1)}{2}$]
Solution:
Formula:
- Here, $1^{2}+2 \cdot 2^{2}+3^{2}+2 \cdot 4^{2}+5^{2}+\ldots$ upto $n$ terms
$ =\frac{n(n+1)^{2}}{2} $
[when $n$ is even] … (i)
When $n$ is odd, $1^{2}+2 \cdot 2^{2}+3^{2}+2 \cdot 4^{2}+5^{2} \ldots+n^{2}$
$ \begin{aligned} & =\{1^{2}+2 \cdot 2^{2}+3^{2}+2 \cdot 4^{2}+\ldots+2(n-1)^{2} \}+n^{2} \\ & =\{\frac{(n-1)(n)^{2}}{2}\}+n^{2} \quad \text { [from Eq. (i)] } \\ & =n^{2} (\frac{n-1}{2}+1)=n^{2} \frac{(n+1)}{2} \end{aligned} $
$\therefore 1^{2}+2 \cdot 2^{2}+3^{2}+2 \cdot 4^{2}+\ldots$ upto $n$ terms, when $n$ is odd
$ =\frac{n^{2}(n+1)}{2} $
BETA
