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Question Number 202780 by BaliramKumar last updated on 03/Jan/24

$$1+2+4+7+11+16+............\mathrm{up}\mathrm{to}25\mathrm{terms}=?$$

Answered by mr W last updated on 03/Jan/24

$$a_1=1$$$$a_2=a_1+1$$$$a_3=a_2+2$$$$a_4=a_3+3$$$$...$$$$a_n=a_{n-1}+(n-1)$$$$\mathrm{\Sigma }:$$$$a_n=1+1+2+3+...+(n-1)=1+\frac{n(n-1)}{2}$$$$S_n=\underset{k=1}{\sum ^n}{a}_k=n+\frac{n(n+1)(2n+1)}{12}-\frac{n(n+1)}{4}$$$$\Rightarrow S_n=\frac{n(n^2+5)}{6}$$$$withn=25:$$$$a_{25}=1+\frac{25\times 24}{2}=301$$$$S_{25}=\frac{25({25}^2+5)}{6}=2625$$$$i.e.$$$$1+2+4+7+11+16+...+301=2625$$

Commented by BaliramKumar last updated on 03/Jan/24

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

Answered by talminator2856792 last updated on 03/Jan/24

$$1+2+3+4+...+n$$$$=\frac{n(n+1)}{2}$$$$$$$$1+3+6+10+...+\frac{n(n+1)}{2}$$$$=\frac{n(n+1)(n+2)}{6}$$$$$$$$1+2+4+7+11+...+a_n$$$$=(1+2+3+4+5+...+n)+$$$$(0+0+1+3+6+...+\frac{(n-1)(n-2)}{2})$$$$=\frac{n(n+1)}{2}+\frac{n(n-1)(n-2)}{6}$$$$$$$$\to 1+2+4+7+11+16+...+a_{25}$$$$=325+2300$$$$=2625$$

Commented by BaliramKumar last updated on 03/Jan/24

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

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