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Mathematical problem of finding the sum of the first n terms of a sequence
Because:12+22+32+...+n 2 = n (n+1) (2n+1)/6.

So:12+22+32+42+...+(n+1) 2 = (n+1) (n+1+1) (2

So: Tn=b 1+b2+b3+...+bn.

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

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

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

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

=-(n+ 1)(n+2)(2n+3)/24+ 1/4

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