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Question Number 28012 by JI Siam last updated on 18/Jan/18

$$1,4,5,16,17,20.......\mathrm{what}\mathrm{is}\mathrm{the}68\mathrm{th}\mathrm{term}\mathrm{in}\mathrm{this}\mathrm{sequnce}?$$

Commented by JI Siam last updated on 18/Jan/18

$$\mathrm{please}\mathrm{give}\mathrm{answer}\mathrm{fast}$$

Answered by ajfour last updated on 18/Jan/18

$$T_{r+4}=2{(-1)}^r-\frac{2}{3}r(r+1)(r+2)+$$$$r(r+1)+7r+14$$$$T_{68}=T_{64+4}$$$$=2-\frac{2}{3}\left(64\times 65\times 66\right)+64\times 65$$$$+7\times 64+14$$$$=2-43\times 64\times 65+448+14$$$$=-43\times 64\times 65+464$$$$=-178416.$$

Commented by ajfour last updated on 19/Jan/18

$$1,4,5,16,17,20,5,...Tseries$$$$3,1,11,1,3,-15,...Sseries$$$$-2,10,-10,2,-18,...Rseries$$$$12,-20,12,-20,...Qseries$$$$-32,32,-32,..Pseries$$$$P_r=32{(-1)}^r$$$$Q_{r+1}-Q_r=P_r$$$$so\underset{r=1}{\sum ^r}(Q_{r+1}-Q_r)=\underset{r=1}{\sum ^r}32{(-1)}^r$$$$\Rightarrow Q_{r+1}-Q_1=16[{(-1)}^r-1]$$$$so{Q}_{r+1}-12=16{(-1)}^r-16$$$$Q_{r+1}=16{(-1)}^r-4$$$$proceedinginasimilarmanner$$$$\underset{r=1}{\sum ^r}(R_{r+2}-R_{r+1})=\underset{r=1}{\sum ^r}{Q}_{r+1}$$$$R_{r+2}=8{(-1)}^r-4r+2$$$$Then\underset{r=1}{\sum ^r}(S_{r+3}-S_{r+2})=\underset{r=1}{\sum ^r}{R}_{r+2}$$$$S_{r+3}=4{(-1)}^r-2r(r+1)+2r+7$$$$andfinally,afteranothersuchstep$$$$T_{r+4}=2{(-1)}^r-\frac{2}{3}r(r+1)(r+2)+$$$$r(r+1)+7r+14.$$

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