The arithmetic progression is grouped into sets of integers, such that the ${{n}^{\mathrm{th}}}$ set

Question

The arithmetic progression is grouped into sets of integers, such that the ${{n}^{\mathrm{th}}}$ set contains $n$ integers as shown. $\left\{ 1 \right\},\left\{ 3,5 \right\},\left\{ 7,9,11 \right\},\left\{ 13,15,17,19 \right\},...\,\,...$ Find the total number of terms in the first $n$ sets, and hence show that the last term of the ${{n}^{\mathrm{th}}}$ set is ${{n}^{2}}+n-1$. [2] Find the first term of the ${{n}^{\mathrm{th}}}$ set. [2] Show that the sum of the terms in the ${{n}^{\mathrm{th}}}$ set is ${{n}^{3}}$. [1] Hence, prove that ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{n}^{3}}=\frac{1}{4}{{n}^{2}}{{\left( n+1 \right)}^{2}}$. [2]

Accepted Answer

(i)(a) $\frac{n(n+1)}{2}$ terms, last term $n^2+n-1$ (b) $n^2-n+1$ (c) $n^3$ (ii) $1^3+2^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}$

2021 CJC Promo Q8

2021 CJC Promo Q8

Video Solution:

Solution:

Solution locked