Question #674b8

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Answer 1 · 2017-01-07T16:14:55

See below.

Calling #S_n=sum_(k=1)^n(-1)^(n-k)k^2# we have

#S_1=1 = (1cdot 2)/2#
#S_2=3=(2 cdot 3)/2#

Supposing that is true for #n#

#S_n=(n(n+1))/2# and knowing that

#S_n -S_(n-1) = n# we have that

#S_(n+1) = (n(n+1))/2+n+1=((n+1)(n+2))/2#

So the summation formula is true.