Find the sum of the following series:

Question:

Find the sum of the following series:

(i) 2 + 5 + 8 + ... + 182

(ii) 101 + 99 + 97 + ... + 47

(iii) (a − b)2 + (a2 + b2) + (b)2 + ... + [(a + b)2 + 6ab]

Solution:

(i) 2 + 5 + 8 + ... + 182

Here, the series is an A.P. where we have the following:

$a=2$

$d=(5-2)=3$

$a_{n}=182$

$\Rightarrow 2+(n-1)(3)=182$

$\Rightarrow 2+3 n-3=182$

$\Rightarrow 3 n-1=182$

$\Rightarrow 3 n=183$

$\Rightarrow n=61$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow S_{61}=\frac{61}{2}[2 \times 2+(61-1) \times 3]$

$=\frac{61}{2}[2 \times 2+60 \times 3]$

$=5612$

(ii) 101 + 99 + 97 + ... + 47

Here, the series is an A.P. where we have the following:

$a=101$

$d=(99-101)=-2$

$a_{n}=47$

$\Rightarrow 101+(n-1)(-2)=47$

$\Rightarrow 101-2 n+2=47$

$\Rightarrow 2 n-2=54$

$\Rightarrow 2 n=56$

$\Rightarrow n=28$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow S_{28}=\frac{28}{2}[2 \times 101+(28-1) \times(-2)]$

$=\frac{28}{2}[2 \times 101+27 \times(-2$

$=2072$

(iii) (a − b)2 + (a2 + b2) + (b)2 + ... + [(a + b)2 + 6ab]

Here, the series is an A.P. where we have the following:

$a=(a-b)^{2}$

$d=\left(a^{2}+b^{2}-(a-b)^{2}\right)=2 a b$

$a_{n}=\left[(\mathrm{a}+\mathrm{b})^{2}+6 a b\right]$

$\Rightarrow(a-b)^{2}+(n-1)(2 a b)=\left[(\mathrm{a}+\mathrm{b})^{2}+6 a b\right]$

$\Rightarrow a^{2}+b^{2}-2 a b+2 a b n-2 a b=\left[a^{2}+b^{2}+2 a b+6 a b\right]$

$\Rightarrow a^{2}+b^{2}-4 a b+2 a b n=a^{2}+b^{2}+8 a b$

$\Rightarrow 2 a b n=12 a b$

$\Rightarrow n=6$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow S_{6}=\frac{6}{2}\left[2(a-b)^{2}+(6-1) 2 a b\right]$

$=3\left[2\left(a^{2}+b^{2}-2 a b\right)+10 a b\right]$

$=3\left[2 a^{2}+2 b^{2}-4 a b+10 a b\right]$

$=3\left[2 a^{2}+2 b^{2}+6 a b\right]$

$=6\left[a^{2}+b^{2}+3 a b\right]$