Sequences And Series Ques 30
- Let $S_k=\frac{1+2+3+\ldots+k}{k}$. If $S_1^2+S_2^2+\ldots+S_{10}^2=\frac{5}{12} $ $A$, then $\mathrm{A}$ is equal to
(2019 Main, 12 Jan I)
(a) $156$
(b) $301$
(c) $283$
(d) $303$
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Answer:
Correct Answer: 30.(d)
Solution: (d) Since, $S_k=\frac{1+2+3+\ldots+k}{k}$
$ =\frac{k(k+1)}{2 k}=\frac{k+1}{2} $
So, $ \quad S_k^2=\left(\frac{k+1}{2}\right)^2=\frac{1}{4}(k+1)^2 $
Now, $\quad \frac{5}{12} A=S_1^2+S_2^2+S_2^2+\ldots S_{10}^2=\sum_{k=1}^{10} S_k^2$
$\Rightarrow \frac{5}{12} A=\frac{1}{4} \sum_{k=1}^{10}(k+1)^2=\frac{1}{4}\left[2^2+3^2+4^2+\ldots 11^2\right]$
$=\frac{1}{4}\left[\frac{11 \times(11+1)(2 \times 11+1)}{6}-1^2\right]$
$\left[\because \sum n^2=\frac{n(n+1)(2 n+1)}{6}\right]$
$=\frac{1}{4}\left[\frac{11 \times 12 \times 23}{6}-1\right]=\frac{1}{4}[(22 \times 23)-1]$
$=\frac{1}{4}[506-1]=\frac{1}{4}[505]$
$\Rightarrow \frac{5}{12} A=\frac{505}{4} \Rightarrow A=303$
BETA
