Answer 1

2870

Given:
∑n = 210

\[\Rightarrow n\left( \frac{n + 1}{2} \right) = 210\]

\[ \Rightarrow n^2 + n - 420 = 0\]

\[ \Rightarrow \left( n - 20 \right)\left( n + 21 \right) = 0\]

\[ \Rightarrow n = 20 \left( \because n > 0 \right)\]

Now,

\[\sum^{}_{} n^2 = \frac{n\left( n + 1 \right)\left( 2n + 1 \right)}{6}\]

\[ \Rightarrow \frac{n(n + 1)}{2} \times \frac{(2n + 1)}{3}\]

\[ \Rightarrow \left( 210 \right) \times \left( \frac{41}{3} \right)\]

\[ \Rightarrow \left( 70 \right) \times \left( 41 \right)\]

\[ \Rightarrow 2870\]