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Question Number 202251 by BaliramKumar last updated on 23/Dec/23

$$\frac{1}{1\times 2\times 3}+\frac{1}{2\times 3\times 4}+\frac{1}{3\times 4\times 5}+..............+\frac{1}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}=?$$

Commented by BaliramKumar last updated on 23/Dec/23

$$\frac{1}{\mathrm{a}(\mathrm{a}+\mathrm{d})(\mathrm{a}+2\mathrm{d})}+\frac{1}{(\mathrm{a}+\mathrm{d})(\mathrm{a}+2\mathrm{d})(\mathrm{a}+3\mathrm{d})}+.......+\frac{1}{\{\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\}\{\mathrm{a}+\mathrm{nd}\}\{\mathrm{a}+(\mathrm{n}+1)\mathrm{d}\}}=?$$

Commented by MATHEMATICSAM last updated on 24/Dec/23

$${\mathrm{T}}_1=\frac{1}{a(a+d)(a+2d)},$$$${\mathrm{T}}_2=\frac{1}{(a+d)(a+2d)(a+3d)},........$$$${\mathrm{T}}_n=\frac{1}{\{a+(n-1)d\}\{a+nd\}\{a+(n+1)d\}}$$$$=\frac{1}{2d}\left[\frac{\{a+(n+1)d\}-\{a+(n-1)d\}}{\{a+(n-1)d\}\{a+nd\}\{a+(n+1)d\}}\right]$$$$=\frac{1}{2d}[\frac{1}{\{a+(n-1)d\}\{a+nd\}}-\frac{1}{\{a+nd\}\{a+(n+1)d\}}]$$$$$$$$$$$${\mathrm{T}}_1=\frac{1}{2d}[\frac{1}{a(a+d)}-\cancel{\frac{1}{(a+d)(a+2d)}}]$$$${\mathrm{T}}_2=\frac{1}{2d}[\cancel{\frac{1}{(a+d)(a+2d)}}-\cancel{\frac{1}{(a+2d)(a+3d)}}]$$$${\mathrm{T}}_3=\frac{1}{2d}[\cancel{\frac{1}{(a+2d)(a+3d)}}-\cancel{\frac{1}{(a+3d)(a+4d)}}]$$$$.$$$$.$$$$.$$$$.$$$${\mathrm{T}}_{n-1}=\frac{1}{2d}[\cancel{\frac{1}{\{a+(n-2)d\}\{a+(n-1)d\}}}-\cancel{\frac{1}{\{a+(n-1)d\}\{a+nd\}}}]$$$${\mathrm{T}}_n=\frac{1}{2d}[\cancel{\frac{1}{\{a+(n-1)d\}\{a+nd\}}}-\frac{1}{\{a+nd\}\{a+(n+1)d\}}]$$$$$$$$\mathrm{Now}{\mathrm{T}}_1+{\mathrm{T}}_2+{\mathrm{T}}_3+....+{\mathrm{T}}_n$$$$=\frac{1}{2d}[\frac{1}{a(a+d)}-\frac{1}{\{a+nd\}\{a+(n+1)d\}}]$$

Commented by BaliramKumar last updated on 24/Dec/23

$$\mathrm{Thanks}\mathrm{Sir}$$$$\mathrm{we}\mathrm{can}\mathrm{also}\mathrm{write}\mathrm{in}\mathrm{this}\mathrm{form}$$$$=\frac{\mathrm{n}}{2}\left[\frac{\mathrm{a}+\{\mathrm{a}+(\mathrm{n}+1)\mathrm{d}\}}{a(a+d)\{\mathrm{a}+\mathrm{nd}\}\{\mathrm{a}+(\mathrm{n}+1)\mathrm{d}\}}\right]$$$$$$

Commented by MATHEMATICSAM last updated on 24/Dec/23

$$\mathrm{yes}$$

Answered by MATHEMATICSAM last updated on 23/Dec/23

$$\mathrm{Let}s=\frac{1}{1\times 2\times 3}+\frac{1}{2\times 3\times 4}+\frac{1}{3\times 4\times 5}+....+\frac{1}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}$$$${\mathrm{T}}_1=\frac{1}{1\times 2\times 3},{\mathrm{T}}_2=\frac{1}{2\times 3\times 4},$$$${\mathrm{T}}_3=\frac{1}{3\times 4\times 5},.......$$$${\mathrm{T}}_n=\frac{1}{n(n+1)(n+2)}=\frac{2}{n(n+1)(n+2)}\times \frac{1}{2}$$$$=\frac{(n+2)-n}{n(n+1)(n+2)}\times \frac{1}{2}=\frac{1}{2}[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}]$$$${\mathrm{T}}_1=\frac{1}{2}[\frac{1}{1\times 2}-\cancel{\frac{1}{2\times 3}}]$$$${\mathrm{T}}_2=\frac{1}{2}[\cancel{\frac{1}{2\times 3}}-\cancel{\frac{1}{3\times 4}}]$$$${\mathrm{T}}_3=\frac{1}{2}[\cancel{\frac{1}{3\times 4}}-\cancel{\frac{1}{4\times 5}}]$$$$.$$$$.$$$$.$$$$.$$$${\mathrm{T}}_{n-1}=\frac{1}{2}[\cancel{\frac{1}{(n-1)n}}-\cancel{\frac{1}{n(n+1)}}]$$$${\mathrm{T}}_n=\frac{1}{2}[\cancel{\frac{1}{n(n+1)}}-\frac{1}{(n+1)(n+2)}]$$$$$$$$\mathrm{Now}$$$$s={\mathrm{T}}_1+{\mathrm{T}}_2+{\mathrm{T}}_3+....+{\mathrm{T}}_n$$$$=\frac{1}{2}[\frac{1}{2}-\frac{1}{(n+1)(n+2)}]$$$$=\frac{1}{2}\left[\frac{(n+1)(n+2)-2}{2(n+1)(n+2)}\right]$$$$=\frac{n^2+3n+2-2}{4(n+1)(n+2)}$$$$=\frac{n(n+3)}{4(n+1)(n+2)}\left(\mathbf{Ans}\right)$$

Commented by BaliramKumar last updated on 23/Dec/23

$$\mathrm{Thanks}\mathrm{Sir}$$

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