#### Question

If a_1=1, a_{n+1}=a_n+n(n+1), find the formula for general terms of the sequence.

Collected in the board: Sequence & Series

Steven Zheng posted 1 day ago

The sequence is a recursive sequence.

Repeat the recursive formula from n to 1

a_n-a_{n-1}=n(n-1)=n^2-n

a{n-1}-a_{n-2}=(n-1)(n-2) =(n-1)^2-(n-1)

\dots

a_3-a_2=3\times 2=3^2-3

a_2-a_1=2\times 1=2^2-2

Addition both sides of all the equations,

a_n=(1^2+2^2+\dots+n^2)-(2+3+\dots+n

=\dfrac{1}{6}n(n+1)(2n+1)-n\cdotp \dfrac{n+1}{2} +1

=\dfrac{n(n-1)(n +1)}{3}+1

Therefore, the formula for general terms of the recursive sequence is

a_n = \dfrac{n(n-1)(n +1)}{3}+1

Steven Zheng posted 1 day ago

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