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Summation and series: YIJC 2024 Promo Q2

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Yishun Innova Junior College2024 PromoPromo Q2●○○ Foundational4 marks

What this question tests

Summation and series.

Question

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

Show full worked solution▾

$\displaystyle\sum_{r=n+1}^{2n}(4r^3+n)=4\sum_{r=n+1}^{2n}r^3+\sum_{r=n+1}^{2n}n$

$=4\left(\sum_{r=1}^{2n}r^3-\sum_{r=1}^{n}r^3\right)+(n)(n)$ $=4\left(\frac{(2n)^2}{4}(2n+1)^2-\frac{n^2}{4}(n+1)^2\right)+n^2$ $=4n^2(2n+1)^2-n^2(n+1)^2+n^2=n^2\left[4(4n^2+4n+1)-(n^2+2n+1)+1\right]$ $=n^2(15n^2+14n+4)$

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