Question

(ii) $S_n=\frac{n^2(n+1{)}^2}{4}$

Answer 1

(ii) Sum of n terms of a sequence, $S_n=\frac{n^2(n+1{)}^2}{4}$

To find the first three terms of the sequence, we need to find S₁, S₂ and S₃.

$$\begin{gathered} \mathrm{For}n=1,\mathrm{we}\mathrm{have}: \\ {S}_1=\frac{1^2(1+1{)}^2}{4}=\frac{1\times 2^2}{4}=\frac{1\times 4}{4}=1 \\ \mathrm{For}n=2,\mathrm{we}\mathrm{have}: \\ {S}_2=\frac{2^2(2+1{)}^2}{4}=\frac{4\times 3^2}{4}=\frac{4\times 9}{4}=9 \\ \mathrm{For}n=3,\mathrm{we}\mathrm{have}: \\ {S}_3=\frac{3^2(3+1{)}^2}{4}=\frac{9\times 4^2}{4}=\frac{9\times 16}{4}=36 \end{gathered}$$

Thus, we get:
t₁ = S₁ = 1
t₂ = S₂ – S₁ = 9 – 1 = 8
t₃ = S₃ – S₂ = 36 – 9 = 27

Therefore, the first three terms of the sequence are 1, 8 and 27.