<ol class="i-list"><li><div class="list-item-paragraph">The arithmetic progression is grouped into sets of integers, such that the ${{n}^{\mathrm{th}}}$ set contains $n$ integers as shown.</div><div class="list-item-paragraph" style="text-align: center">$\left\{ 1 \right\},\left\{ 3,5 \right\},\left\{ 7,9,11 \right\},\left\{ 13,15,17,19 \right\},...\,\,...$</div><ol class="abc-list"><li><div class="list-item-paragraph">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 ${{n}^{2}}+n-1$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find the first term of the ${{n}^{\mathrm{th}}}$ set.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Show that the sum of the terms in the ${{n}^{\mathrm{th}}}$ set is ${{n}^{3}}$.</div><p style="text-align: right;">[1]</p></li></ol></li><li><div class="list-item-paragraph">Hence, prove that ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{n}^{3}}=\frac{1}{4}{{n}^{2}}{{\left( n+1 \right)}^{2}}$.</div><p style="text-align: right;">[2]</p></li></ol>

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

$\left\{ 1 \right\},\left\{ 3,5 \right\},\left\{ 7,9,11 \right\},\left\{ 13,15,17,19 \right\},...\,\,...$

Hence, prove that ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{n}^{3}}=\frac{1}{4}{{n}^{2}}{{\left( n+1 \right)}^{2}}$.

2021 CJC Promo Q8

2021 CJC Promo Q8

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