Question #b5167

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Answer 1 · 2016-09-03T23:15:54

See below.

$1/4 + 9/4 + 18/4+ ... [2(n-1)+1/4] = (4n^2-3n)/4$

For $n = 1$ we have $1/4 = (4 xx 1^2-3 xx 1)/4 = 1/4$

Supposing that is true for $n$

$sum_(k=1)^n (2(k-1)+1/4) = (4n^2-3n)/4$

then verify the correctness for $n+1$

$sum_(k=1)^(n+1) (2(k-1)+1/4) = sum_(k=1)^n (2(k-1)+1/4) + 2n+1/4$

$(4n^2-3n)/4 + 2n+1/4 = (4(n+1)^2-3(n+1))/4$

So we conclude that is is true for $n+1$ and by mathematical induction, the assertion is true.