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Question Number 64126 by Cheyboy last updated on 13/Jul/19

Answered by ajfour last updated on 13/Jul/19

$$320-8-10-24$$$$-32-8-2-14$$$$246-12$$$$-18-18$$$$d_n=-18$$$$c_n=24-18(n-1)$$$$b_{n+1}-b_n=c_n$$$$\underset{n=1}{\sum ^n}(b_{n+1}-b_n)=\underset{n=1}{\sum ^n}{c}_n$$$$\Rightarrow b_{n+1}-b_1=24n-9n(n-1)$$$$\Rightarrow b_{n+1}=33n-9n^2-32$$$$\underset{n=1}{\sum ^n}(a_{n+2}-a_{n+1})=\underset{n=1}{\sum ^n}{b}_{n+1}$$$$a_{n+2}-a_2=\frac{33n(n+1)}{2}-\frac{9n(n+1)(2n+1)}{6}-32n$$$$a_n=\frac{33(n-2)(n-1)}{2}-\frac{3(n-2)(n-1)(2n-3)}{2}-32(n-2)$$$$=3(n-1)(n-2)(7-n)-32(n-2)$$$$=3(-n^3+10n^2-23n+14)-32n+64$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$a_n=-3n^3+30n^2-101n+106$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$

Commented by Cheyboy last updated on 13/Jul/19

$$Thankyousir$$

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