Question

Question #b5167

Answer 1

See below.

Explanation:

`1/4 + 9/4 + 18/4+ ... [2(n-1)+1/4] = (4n^2-3n)/4`

For `n = 1` we have `1/4 = (4 xx 1^2-3 xx 1)/4 = 1/4`

Supposing that is true for `n`

`sum_(k=1)^n (2(k-1)+1/4) = (4n^2-3n)/4`

then verify the correctness for `n+1`

`sum_(k=1)^(n+1) (2(k-1)+1/4) = sum_(k=1)^n (2(k-1)+1/4) + 2n+1/4`

`(4n^2-3n)/4 + 2n+1/4 = (4(n+1)^2-3(n+1))/4`

So we conclude that is is true for `n+1` and by mathematical induction, the assertion is true.