Question

1+2+3+....+ n < 1/8 (2n +1) square

Answer 1

here's your answer
hope it helps

Answer 2

Hope u like my process
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=> sum of the series

$1 + 2 + 3 + 4 + .... + n = \frac{n(n + 1)}{2}$

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Now,
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$= > \frac{(2n + 1) ^{2} }{8} \\ \\ = \frac{4 {n}^{2} + 4n + 1}{8} \\ \\ = \frac{4 {n}^{2} + 4n }{8} + \frac{1}{8} \\ \\ = \frac{4n(n + 1)}{8} + \frac{1}{8} \\ \\ = \frac{n(n + 1)}{2} + \frac{1}{8} \\ \\ = 1 + 2 + 3 + 4 + ... + n + \frac{1}{8}$


Thus now..
It can be proved that..

$1 + 2 + 3 + 4 + .... + n < \frac{ {(2n + 1)}^{2} }{8}$
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Hope this is ur required answer

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