Question

Find the sum of the following series:
(i) 2 + 5 + 8 + ... + 182
(ii) 101 + 99 + 97 + ... + 47
(iii) (a − b)² + (a² + b²) + (a + b)² + ... + [(a + b)² + 6ab]

Answer 1

(i) 2 + 5 + 8 + ... + 182
Here, the series is an A.P. where we have the following:
$$\begin{gathered} a=2 \\ d=\left(5-2\right)=3 \\ {a}_n=182 \\ \Rightarrow 2+(n-1)\left(3\right)=182 \\ \Rightarrow 2+3n-3=182 \\ \Rightarrow 3n-1=182 \\ \Rightarrow 3n=183 \\ \Rightarrow n=61 \\ {S}_n=\frac{n}{2}\left[2a+(n-1)d\right] \\ \Rightarrow S_{61}=\frac{61}{2}\left[2\times 2+\left(61-1\right)\times 3\right] \\ =\frac{61}{2}\left[2\times 2+60\times 3\right] \\ =5612 \end{gathered}$$

(ii) 101 + 99 + 97 + ... + 47
Here, the series is an A.P. where we have the following:
$$\begin{gathered} a=101 \\ d=\left(99-101\right)=-2 \\ {a}_n=47 \\ \Rightarrow 101+(n-1)(-2)=47 \\ \Rightarrow 101-2n+2=47 \\ \Rightarrow 2n-2=54 \\ \Rightarrow 2n=56 \\ \Rightarrow n=28 \\ {S}_n=\frac{n}{2}\left[2a+(n-1)d\right] \\ \Rightarrow S_{28}=\frac{28}{2}\left[2\times 101+\left(28-1\right)\times (-2)\right] \\ =\frac{28}{2}\left[2\times 101+27\times (-2)\right] \\ =2072 \end{gathered}$$

(iii) (a − b)² + (a² + b²) + (a + b)² + ... + [(a + b)² + 6ab]
Here, the series is an A.P. where we have the following:
$$\begin{gathered} a=(a-b{)}^2 \\ d=\left(a^2+b^2-(a-b{)}^2\right)=2ab \\ {a}_n={[(a\; +\; b)}^2+6ab] \\ \Rightarrow (a-b{)}^2+(n-1\left)\right(2ab)={[(a\; +\; b)}^2+6ab] \\ \Rightarrow a^2+b^2-2ab+2abn-2ab=[a^2+b^2+2ab+6ab] \\ \Rightarrow a^2+b^2-4ab+2abn=a^2+b^2+8ab \\ \Rightarrow 2abn=12ab \\ \Rightarrow n=6 \\ {S}_n=\frac{n}{2}\left[2a+(n-1)d\right] \\ \Rightarrow S_6=\frac{6}{2}\left[2(a-b{)}^2+\left(6-1\right)2ab\right] \\ =3\left[2(a^2+b^2-2ab)+10ab\right] \\ =3\left[2a^2+2b^2-4ab+10ab\right] \\ =3\left[2a^2+2b^2+6ab\right] \\ =6\left[a^2+b^2+3ab\right] \end{gathered}$$