Answer to Question #186556 in Discrete Mathematics for Maricel

Question #186556

Solve the following. (10 pts each)

1. Prove P(n) = n2(n + 1)

2. Recurrence relation an = 2n with the initial term a1= 2.


1
Expert's answer
2021-05-07T09:47:39-0400

(1) P(n)=n2(n+1)P(n)=n^2(n+1)\\

=n3+n= n^3+n \\


Sn=P(n)=n3+nS_n= \sum P(n)=\sum n^3+\sum n

=(n(n+1)2)2+n(n+1)2=n4+n3+2n2+n3+n2+2n4=14n(n+1)(n2+n+2)= (\dfrac{n(n+1)}{2})^2 + \dfrac{n(n+1)}{2}\\ \\ =\dfrac{n^4+n^3+2n^2+n^3+n^2+2n}{4}\\=\dfrac{1}{4}n(n+1)(n^2+n+2)


(2) an=2n      a1=2a_n=2n \ \ \ \ \ \ a_1=2

This is already a recurrence relation with sum

Sn=2n=2n(n+1)2=n(n+1)S_n=\sum 2n=2\dfrac{n(n+1)}{2}=n(n+1)


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