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Question Number 185542 by mathlove last updated on 23/Jan/23

Answered by Ar Brandon last updated on 23/Jan/23

$$u_n=3,6,10,15,21,...d_0=3$$$$\mathrm{\Delta }{u}_n=3,4,5,6,...d_1=3$$$${\mathrm{\Delta }}^2{u}_n=1,1,1,...d_2=1$$$$u_n=d_0+\frac{d_1(n-1)}{1!}+\frac{d_2(n-1)(n-2)}{2!}$$$$=3+3(n-1)+\frac{(n-1)(n-2)}{2}=\frac{n^2}{2}+\frac{3n}{2}+1$$$$S=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\cdot \cdot \cdot =\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{2}{n^2+3n+2}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{2}{(n+2)(n+1)}=\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{2}{n+1}-\frac{2}{n+2})$$$$=2-\underset{n\to \mathrm{\infty }}{\lim }\frac{2}{n+2}=2$$

Commented by mathlove last updated on 23/Jan/23

$$descraibtosimple$$$$Howis{u}_n=3,6,10,15,21.....d_0=3{what}^{\prime }s{d}_0$$$$andanutherone??$$

Commented by Ar Brandon last updated on 23/Jan/23

$$u_n\mathrm{is}\mathrm{obtained}\mathrm{using}\mathrm{the}\mathrm{terms}\mathrm{at}\mathrm{the}\mathrm{denominator}.$$$$d_0\mathrm{is}\mathrm{the}\mathrm{first}\mathrm{term}.$$

Answered by Frix last updated on 23/Jan/23

$$a_n=\frac{2}{n(n+1)}$$$$\underset{k=1}{\sum ^n}{a}_k=2-\frac{2}{n+1}$$$$\Rightarrow \mathrm{answer}\mathrm{is}2$$

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