Question

What is the sum of the series *picture attached*

Answer 1

Answer:

$\sum_{n=4}^{18}(n+1)^2=2440$

Step-by-step explanation:

Series: $\sum_{n=4}^{18} (n+1)^2$

Formula :$\sum_{k=m}^n=\sum_{k=1}^n-\sum_{k=1}^{m-1}$

$\Rightarrow \sum_{n=4}^{18}(n+1)^2=\sum_{n=1}^{18}(n+1)^2-\sum_{n=1}^{3}(n+1)^2$

$\sum_{n=1}^{18}(n+1)^2=2469$

$\sum_{n=1}^{3}(n+1)^2=29$

$\Rightarrow \sum_{n=4}^{18}(n+1)^2=2469-29$

$\Rightarrow \sum_{n=4}^{18}(n+1)^2=2440$

So, Option C is true