Prove the following by using the principle of mathematical induction for all `n in N`:`1 + 2 + 3 + ... + n <1/8(2n+1)^2`.

Explanation

`p(n) = 1 + 2 + 3 + ....+n < 1/8(2n+1)^2`

`p(1) = 1 < 1/8(2+1)^2 = 9/8`

`1< 9/8`so, p(1) is true

`p(k) = 1+ 2+ 3 + ....+ k < 1/8(2k+1)^2`

`p(k+1) = 1+2 +3 +...+k + k+1 < 1/8(2k+1) + k+1 < 1/8[(2k+1)^2 + 8(k+1)] < 1/8(4k^2 + 4k + 1 + 8k +8) < 1/8 (4k^2 + 12k + 9)`

`< 1/8 (2k+3)^2`

`< 1/8[2(k+1) + 1]^2`

`p(k+1) `is ture whenever p(k) is true

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