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Worked solution

Express consecutive pair products using square and linear sums

Given ∑r=1nr2=n(n+1)(2n+1)/6\sum_{r=1}^nr^2=n(n+1)(2n+1)/6, find the sum of the first nn terms of (1)(2)+(3)(4)+(5)(6)+⋯(1)(2)+(3)(4)+(5)(6)+\cdots. marks

[2018/MI/Promo/10(b)]

CJC · Review problems · Question 5

Answer

n(n+1)(4n−1)/3n(n+1)(4n-1)/3.

Identify and expand the r-th product

The r-th term is (2r−1)(2r)=4r2−2r(2r-1)(2r)=4r^2-2r. Hence the required sum is 4∑r=1nr2−2∑r=1nr4\sum_{r=1}^nr^2-2\sum_{r=1}^nr.

How did this part go?

Substitute and factor the common expression

S=23n(n+1)(2n+1)−n(n+1)=n(n+1)[2(2n+1)−3]/3=n(n+1)(4n−1)/3S=\tfrac23n(n+1)(2n+1)-n(n+1)=n(n+1)[2(2n+1)-3]/3=n(n+1)(4n-1)/3.

How did this part go?

Common mistakes

The odd and even factors belong to the same term; they cannot be summed independently and multiplied.

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