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Summation and series: YIJC 2024 Promo Q2

Yishun Innova Junior College2024 PromoPromo Q2●○○ Foundational4 marks

What this question tests

Summation and series.

Question

Given that ∑r=1nr3=n24(n+1)2\displaystyle\sum_{r=1}^{n}r^3=\frac{n^2}{4}(n+1)^2, find an expression in terms of nn for ∑r=n+12n(4r3+n)\displaystyle\sum_{r=n+1}^{2n}(4r^3+n). Leave your answer in fully factorised form. [4]

Show full worked solution▾

∑r=n+12n(4r3+n)=4∑r=n+12nr3+∑r=n+12nn\displaystyle\sum_{r=n+1}^{2n}(4r^3+n)=4\sum_{r=n+1}^{2n}r^3+\sum_{r=n+1}^{2n}n

=4(∑r=12nr3−∑r=1nr3)+(n)(n)=4\left(\sum_{r=1}^{2n}r^3-\sum_{r=1}^{n}r^3\right)+(n)(n) =4((2n)24(2n+1)2−n24(n+1)2)+n2=4\left(\frac{(2n)^2}{4}(2n+1)^2-\frac{n^2}{4}(n+1)^2\right)+n^2 =4n2(2n+1)2−n2(n+1)2+n2=n2[4(4n2+4n+1)−(n2+2n+1)+1]=4n^2(2n+1)^2-n^2(n+1)^2+n^2=n^2\left[4(4n^2+4n+1)-(n^2+2n+1)+1\right] =n2(15n2+14n+4)=n^2(15n^2+14n+4)

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