Question

Show that:

$\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+....+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$

Answer 1

consider the given series.

$\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+.....+\frac{1}{n(n+1)(n+2)}$

$r^{th}term,t_r=\frac{1\times 2}{2r(r+1(r+2)}=\frac{(r+2)-r}{2r(r+1)(r+2)}$

$=\frac{1}{2}[\frac{(r+2)}{r(r+1)(r+2)}-\frac{r}{r(r+1)(r+2)}]$

$=\frac{1}{2}[\frac{1}{r(r+1)}-\frac{1}{(r+1)(r+2)}]$

Put $r=1$, we get

$t_1=\frac{1}{2}[\frac{1}{1.2}-\frac{1}{2.3}]$

Put $r=2$, we get

$t_2=\frac{1}{2}[\frac{1}{2.3}-\frac{1}{3.4}]$

Put $r=3$, we get

$t_3=\frac{1}{2}[\frac{1}{3.4}-\frac{1}{4.5}]$

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Put $r=n$, we get

$t_n=\frac{1}{2}[\frac{1}{n(n+1)}-\frac{1}{(n+1(n+2)}]$

Therefore, the sum of this series,

$S_n={\sum }_{r=1}^n{t}_r$

$S_n=\frac{1}{2}[\frac{1}{1.2}-\frac{1}{(n+1(n+2)}]$

$S_n=\frac{1}{4(n+1)(n+2)}\left[\right(n+1\left)\right(n+2)-2]$

$S_n=\frac{1}{4(n+1)(n+2)}[n^2+3n+2-2]$

$S_n=\frac{n(n+3)}{4(n+1)(n+2)}$

Since,

$LHS=RHS$

Hence proved.