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.