Question : If $\small 1^{2}+2^{2}+3^{2}+......+\ p^{2}=\frac{p(p+1)(2p+1)}{6}$, then $\small 1^{2}+3^{2}+5^{2}+......+17^{2}$ is equal to:
Option 1: 1785
Option 2: 1700
Option 3: 980
Option 4: 969
Correct Answer: 969
Solution :
Given: $1^{2}+2^{2}+3^{2}+......+p^{2}=\frac{p(p+1)(2p+1)}{6}$
So, $1^{2}+2^{2}+3^{2}+......+17^{2}=\frac{17(17+1)(2\times 17+1)}{6}$-----------------(1)
Also, $2^{2}+4^{2}+6^{2}+......+16^{2}=2^{2}(1^{2}+2^{2}+3^{2}+......+8^{2})$
$⇒2^{2}+4^{2}+6^{2}+......+16^{2}=2^{2}(\frac{8(8+1)(2\times 8+1)}{6})$----------------(2)
Subtracting equation (2) from equation (1), we get,
$1^{2}+3^{2}+5^{2}+......+17^{2}=\frac{17(17+1)(2\times 17+1)}{6}-2^{2}(\frac{8(8+1)(2\times 8+1)}{6})$
$\therefore1^{2}+3^{2}+5^{2}+......+17^{2}= 969$
Hence, the correct answer is 969.
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