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Prove the following by using the principle of mathematical induction for all n ∈ N: \[\] $1+2+3+...+n<\large\frac{1}{8}$$(2n+1)^2$

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Let the given statement be $P(n)$, i.e.,
$P(n) :1+2+3+...+n<\large\frac{1}{8}$$(2n+1)^2$
It can be noted that P(n) is true for $n=1$ since $1 < \large\frac{1}{8}$$(2.1+1)^2=\large\frac{9}{8}$.
Let $P(k)$ be true for some positive integer $k$, i.e.,
We shall now prove that $P(k+1)$ is true whenever $P(k)$ is true.
$(1+2+...+k)+(k+1)< \large\frac{1}{8}$$(2k+1)^2+(k+1)\qquad$ [ Using (1)]
$< \large\frac{1}{8}$$\{ (2k+1)^2+8(k+1) \}$
$< \large\frac{1}{8}$$\{4k^2+4k+1+8k+8\}$
$< \large\frac{1}{8}$$\{4k^2+12k+9\}$
$< \large\frac{1}{8}$$(2k+3)^2$
$< \large\frac{1}{8}$$\{2(k+1)+1\}^2$
Hence $(1+2+3+...+k)+(k+1)< \large\frac{1}{8}$$(2k+1)^2+(k+1)$
Thus, $P(k+1)$ is true whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., $n$.
answered May 1, 2014 by thanvigandhi_1
edited May 1, 2014 by thanvigandhi_1

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