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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}$$(2n+1)^2 Can you answer this question? ## 1 Answer 0 votes Let the given statement be P(n), i.e., P(n) :1+2+3+...+n<\large\frac{1}{8}$$(2n+1)^2$
It can be noted that P(n) is true for $n=1$ since $1 < \large\frac{1}{8}$$(2.1+1)^2=\large\frac{9}{8}. Let P(k) be true for some positive integer k, i.e., 1+2+...+k<\large\frac{1}{8}$$(2k+1)^2$----------(1)
We shall now prove that $P(k+1)$ is true whenever $P(k)$ is true.
Consider
$(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$