Prove by induction, 2? Thanks :)

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1 Answer
Mar 25, 2018

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

sum_{i=1}^n (4i-2) = 2n^2ni=1(4i2)=2n2

is true. Then

sum_{i=1}^{n+1} (4i-2) =[sum_{i=1}^{n} (4i-2)]+4(n+1)-2 n+1i=1(4i2)=[ni=1(4i2)]+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.