Porisma

Worked solution

Combine polynomial and geometric finite sums

Given ∑r=1nr2=n(n+1)(2n+1)6\sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}6 and ∑r=1nr3=[n2(n+1)]2\sum_{r=1}^{n}r^3=[\frac n2(n+1)]^2, evaluate: (a) 1(1)+2(3)+3(5)+⋯+n(2n−1)1(1)+2(3)+3(5)+\cdots+n(2n-1); (b) ∑r=0n(ar3+ar−a)\sum_{r=0}^{n}(ar^3+a^r-a) in terms of aa (and nn); (c) (n+1)2+(n+2)2+⋯+(2n)2(n+1)^2+(n+2)^2+\cdots+(2n)^2.

CJC · Sequences and Series · Example 29

(a) Products with consecutive odd numbers

Answer

n(n+1)(4n−1)6\frac{n(n+1)(4n-1)}6.

Express the products with one index

∑r=1nr(2r−1)=2∑r=1nr2−∑r=1nr\begin{aligned}&\sum_{r=1}^{n}r(2r-1)\\ & =2\sum_{r=1}^{n}r^2-\sum_{r=1}^{n}r\end{aligned}

The polynomial expansion separates a square sum and a natural-number sum.

Substitute and factor

2n(n+1)(2n+1)6−n(n+1)2=n(n+1)6(4n+2−3)=n(n+1)(4n−1)6.\begin{aligned}\frac{2n(n+1)(2n+1)}6-\frac{n(n+1)}2&=\frac{n(n+1)}6(4n+2-3)\\&=\frac{n(n+1)(4n-1)}6.\end{aligned}

Retain the source intermediate factor before the final simplification.

How did this part go?

(b) Cubes, powers and a repeated constant

Answer

For a≠1a\ne1: an2(n+1)24+1−an+11−a−(n+1)a\frac{an^2(n+1)^2}{4}+\frac{1-a^{n+1}}{1-a}-(n+1)a. For a=1a=1: n2(n+1)24\frac{n^2(n+1)^2}{4}.

Separate all three sums and retain the zero lower limit

∑r=0n(ar3+ar−a)=a∑r=0nr3+∑r=0nar−∑r=0na=a∑r=1nr3+(1+a+a2+⋯+an)−(n+1)a.\begin{aligned}\sum_{r=0}^{n}(ar^3+a^r-a)&=a\sum_{r=0}^{n}r^3+\sum_{r=0}^{n}a^r-\sum_{r=0}^{n}a\\&=a\sum_{r=1}^{n}r^3+(1+a+a^2+\cdots+a^n)-(n+1)a.\end{aligned}

The cube at r=0 adds zero; the geometric and constant sums contain n+1 terms.

Use the cube formula and the finite GP formula

an2(n+1)24+1−an+11−a−(n+1)a(a≠1)\frac{an^2(n+1)^2}{4}+\frac{1-a^{n+1}}{1-a}-(n+1)a\qquad(a\ne1)

The powers start with 1 and have ratio a. The printed quotient assumes a≠1; it is a finite sum and has no convergence restriction.

Handle the missing unit-ratio case

a=1:∑r=0n(r3+1−1)=∑r=1nr3=n2(n+1)24a=1:\quad\sum_{r=0}^{n}(r^3+1-1)=\sum_{r=1}^{n}r^3=\frac{n^2(n+1)^2}{4}

The prompt does not exclude a=1. Then each summand is r³ because the added 1 and subtracted 1 cancel.

How did this part go?

(c) The next n square terms

Answer

n6(2n+1)(7n+1)\frac n6(2n+1)(7n+1).

Subtract the first n squares from the first 2n squares

∑r=n+12nr2=∑r=12nr2−∑r=1nr2\begin{aligned}&\sum_{r=n+1}^{2n}r^2\\ & =\sum_{r=1}^{2n}r^2-\sum_{r=1}^{n}r^2\end{aligned}

The desired interval starts at n+1, so subtract through n, not n−1.

Substitute and simplify

2n(2n+1)(4n+1)6−n(n+1)(2n+1)6=n6(2n+1)[8n+2−(n+1)]=n6(2n+1)(7n+1).\begin{aligned}\frac{2n(2n+1)(4n+1)}6-\frac{n(n+1)(2n+1)}6&=\frac n6(2n+1)[8n+2-(n+1)]\\&=\frac n6(2n+1)(7n+1).\end{aligned}

This retains the source expansion and factorisation.

How did this part go?

Solution notes

The source says part (b) is in terms of a; n remains a free upper bound, so the result depends on both. Its printed quotient omits the a=1 case. At a=0, interpret 1+a+⋯+a^n as the finite polynomial with constant term 1; this gives 1 without assigning an independent analytic meaning to 0^0.