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Question Number 37692 by kunal1234523 last updated on 16/Jun/18

$$2+6+12+20+30+42+.........+\mathrm{n}=\frac{1}{3}\left(\mathrm{n}\right)(\mathrm{n}-1)(\mathrm{n}-2)$$$$\mathrm{is}\mathrm{this}\mathrm{true}.\mathrm{if}\mathrm{yes}\mathrm{so}\mathrm{please}\mathrm{derive}\mathrm{it}\mathrm{from}\mathrm{L}.\mathrm{H}.\mathrm{S}$$$$\mathrm{hey}\mathrm{I}\mathrm{just}\mathrm{noticed}\mathrm{I}\mathrm{got}15\mathrm{yrs}\mathrm{and}5\mathrm{months}\mathrm{older}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jun/18

$$s=2+6+12+20+30+42+...+T_{n-1}+T_n$$$$s=2+6+12+20+......+T_{n-2}+T_{n-1}+T_n$$$$shiftingrightandsubstructing$$$$0=2+4+6+8+10+...+(-T_n)$$$$T_n=2+4+6+8...uptonterms$$$$T_n=\frac{n}{2}[2\times 2+(n-1)\times 2$$$$T_n=\frac{n}{2}[4+2n-2]$$$$T_n=\frac{n}{2}[2n+2]$$$$T_n=n^2+n$$$$s=\underset{1}{\sum ^n}{n}^2+\underset{1}{\sum ^n}n$$$$s=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$$$$s=\frac{n(n+1)}{2}(\frac{2n+1}{3}+1)$$$$s=\frac{n(n+1)}{2}\left(\frac{2n+4}{3}\right)$$$$=\frac{n(n+1)(n+2)}{3}$$$$$$$$$$

Commented by kunal1234523 last updated on 17/Jun/18

$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{sir}$$

Answered by ajfour last updated on 16/Jun/18

$$2612203042n{S}_{series}$$$$4681012T_{series}$$$$T_n=2+2n$$$$S_{n+1}-S_n=T_n$$$$\mathrm{\Sigma }(S_{n+1}-S_n)=\mathrm{\Sigma }{T}_n$$$$S_{n+1}-S_1=\mathrm{\Sigma }(2+2n)$$$$S_{n+1}-2=2n+n(n+1)$$$$S_n=2+2(n-1)+(n-1)n$$$$=n^2+n=n(n+1)$$$$\mathrm{\Sigma }{S}_n=\mathrm{\Sigma }n(n+1)$$$$=\frac{n(n+1)(n+2)}{3}.$$

Commented by kunal1234523 last updated on 17/Jun/18

$$\mathrm{thanks}\mathrm{a}\mathrm{lot}$$

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