Question

Let $A$ be the sum of the first $20$ terms and $B$ be the sum of the first $40$ terms of the series $1$ + ${2.2}^2$ + $3^2$ + ${2.4}^2$ + $5^2+{2.6}^2$ +......... Find the value of $A$.

• A. $496$
• B. $232$
• C. $248$
• D. $464$

Answer 1

The correct option is A $496$

Sum of first $20$ terms -

$\Rightarrow (1^2+3^2+\_\_\_\_+{19}^2)+2(2^2+4^2+\_\_\_\_+{20}^2)$

$\Rightarrow A=(1^2+2^2+3^2\_\_\_\_+{19}^2+{20}^2)+(2^2+4^2+\_\_\_\_+{20}^2)$

$\Rightarrow A=\frac{20\times 41\times 21}{6}+1^2(1^2+2^2+\_\_\_\_+{10}^2)$

$\Rightarrow A=2870+4\times \frac{10\times 11\times 21}{6}$

$\Rightarrow A=2870+1540$

$\Rightarrow A=4410.$

Hence, the answer is $4410.$