Question

Solve:
$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+---+\frac{1}{(3n-1)(3n+2)}$

Answer 1

Let $S\left(n\right)=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+.........+\frac{1}{(3n-1)(3n-2)}$

Step - (1)

prove s(n) for n=1

$L.H.S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+.......+\frac{1}{\left(3\right(1)-1)\left(3\right(1)+2)}$

$=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+........+\frac{1}{2.5}$

$L.H.S=\frac{1}{2.5}=\frac{1}{10}$

$R.H.S=\frac{1}{2.5}=\frac{1}{10}$

So, it is true for n=1.

Step (2)

Assume that S(n) is true for some n(k)

$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+........+\frac{1}{(3k-1)(3k+2)}$

$=\frac{k}{(6k+4)}$

Step - (3)

Prove s(n) for n= k+1

$L.H.S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+......+\frac{1}{\left(3\right(k+1)-1)\left(3\right(k+1)+2)}$

$=\frac{k}{(6k+4)}+\frac{1}{(3k+2)(3k+5)}$

$=\frac{k}{2(3k+2)}+\frac{1}{(3k+2)(3k+5)}$

$=[3k^2+5k+2]/\left[2\right(3k+2\left)\right(3k+5\left)\right]$

$\left[\right(3k+2\left)\right(k+1\left)\right]/\left[2\right(3k+2\left)\right(3k+5\left)\right]$

$=\frac{k+1}{2(3k+5)}$

$=\frac{k+1}{6k+10}$