Prove by induction, 2? Thanks :)

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Prove by induction, 2? Thanks :)

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Answer 1 · 2018-03-25T18:57:29

Let us assume that the statement is true for some $n$ $n$ , i.e, for some $n$ $n$ the relation

$n \sum i = 1 (4 i - 2) = 2 n^{2}$ $sum_{i=1}^n (4i-2) = 2n^2$

is true. Then

$n + 1 \sum i = 1 (4 i - 2) = [n \sum i = 1 (4 i - 2)] + 4 (n + 1) - 2$ $sum_{i=1}^{n+1} (4i-2) =[sum_{i=1}^{n} (4i-2)]+4(n+1)-2$
$qquad = 2n^2+4n+2 = 2(n^2+2n+1) = 2(n+1)^2$

Thus, if the statement is true for $n$ , it is true for $n+1$ .

Now, the statement is obviously true for $n=1$ . Hence, by mathematical induction, it is true for all natural numbers $n$ .

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Prove by induction, 2? Thanks :)

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