Question

`(1)+(1+3)+(1+3+5)+......+(1+3+5+...+21)`?

Answer 1

Sum is `506`

Explanation:

The given series `(1)+(1+3)+(1+3+5)+....+(1+3+5+...+21)` is equivalent to

`1+4+9+.....+121` or `1^2+2^2+3^2+...+11^2`

hence, this is sum up to `11^(th)` term of series `sumn^2`.

Sum up to `n` terms of series `sumn^2=(n(n+1)(2n+1))/6`

hence desired sum is `(11xx12xx23)/6=506`

However to prove `sumn^2=(n(n+1)(2n+1))/6`, one can use induction method or a much more laborious proof is given here.