Series - Problem 1

$1\times1^2 +2\times3^2 +3\times5^2 +..........+50\times99^2$

$$ $\displaystyle = \sum_{n=1}^{50} n(2n-1)^2$

$$ $\displaystyle = \sum_{n=1}^{50} 4n^3-4n^2+n$

$$ $\displaystyle = \sum_{n=1}^{50} 4n^3-\sum_{n=1}^{50}4n^2+\sum_{n=1}^{50}n$

$$

$\displaystyle \sum 4n^3 = 4 \times \frac{n^2(n+1)^2}{4} =n^2(n+1)^2$

$$ $\displaystyle \sum 4n^2 = 4 \times \frac{n(n+1)(2n+1)}{6} =\frac{2n(n+1)(2n+1)}{3}$

$$ $\displaystyle \sum n = \frac{n(n+1)}{2}$

$$ Now,

$\displaystyle \sum_{n=1}^{50} 4n^3-\sum_{n=1}^{50}4n^2+\sum_{n=1}^{50}n\\ = 50^2\times 51^2 - \frac{2\times50\times51\times101}{3}+\frac{50\times51}{2}\\ =\boxed{6332075}$

$Sum\quad to\quad the\quad { n }^{ th }\quad term\quad is\quad given\quad by\quad =\quad \frac { n{ (n+1) }({ 6n }^{ 2 }-2n-1) }{ 6 }$