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Question Number 148677 by nadovic last updated on 30/Jul/21

  A series of natural numbers are     grouped as 1+(2+3)+(4+5+6)+...    such that the rth group contains r     terms. Show that the sum of the    numbers in the (2r−1)th group is                          r^4 −(r−1)^4 .

$$\:\:\mathrm{A}\:\mathrm{series}\:\mathrm{of}\:\mathrm{natural}\:\mathrm{numbers}\:\mathrm{are}\: \\ $$$$\:\:\mathrm{grouped}\:\mathrm{as}\:\mathrm{1}+\left(\mathrm{2}+\mathrm{3}\right)+\left(\mathrm{4}+\mathrm{5}+\mathrm{6}\right)+... \\ $$$$\:\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:{rth}\:\mathrm{group}\:\mathrm{contains}\:{r}\: \\ $$$$\:\:\mathrm{terms}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the} \\ $$$$\:\:\mathrm{numbers}\:\mathrm{in}\:\mathrm{the}\:\left(\mathrm{2}{r}−\mathrm{1}\right){th}\:\mathrm{group}\:\mathrm{is} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{r}}^{\mathrm{4}} −\left(\boldsymbol{{r}}−\mathrm{1}\right)^{\mathrm{4}} . \\ $$

Answered by gsk2684 last updated on 30/Jul/21

t_1 =1  t_2 =((1)+1)+((1)+2)  t_3 =((1+2)+1)+((1+2)+2)+((1+2)+3)  ....  t_(2r−1) =  ((1+2+3+..+(2r−2)+1)+  ((1+2+3+..+(2r−2)+2)+  ((1+2+3+..+(2r−2)+3)+...  ((1+2+3+..+(2r−2)+2r−1)  =(2r−1)(1+2+3+..+(2r−2))+(1+2+3+..+(2r−1))  =(2r−1)(((2r−2)(2r−1))/2)+(((2r−1)(2r))/2)  =(r−1)(2r−1)^2 +(2r−1)r  =(2r−1){2r^2 −3r+1+r}  =(2r−1)(2r^2 −2r+1)    r^4 −(r−1)^4   =(r^2 −(r−1)^2 )(r^2 +(r−1)^2 )  =(2r−1)(2r^2 −2r+1)

$${t}_{\mathrm{1}} =\mathrm{1} \\ $$$${t}_{\mathrm{2}} =\left(\left(\mathrm{1}\right)+\mathrm{1}\right)+\left(\left(\mathrm{1}\right)+\mathrm{2}\right) \\ $$$${t}_{\mathrm{3}} =\left(\left(\mathrm{1}+\mathrm{2}\right)+\mathrm{1}\right)+\left(\left(\mathrm{1}+\mathrm{2}\right)+\mathrm{2}\right)+\left(\left(\mathrm{1}+\mathrm{2}\right)+\mathrm{3}\right) \\ $$$$.... \\ $$$${t}_{\mathrm{2}{r}−\mathrm{1}} = \\ $$$$\left(\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+..+\left(\mathrm{2}{r}−\mathrm{2}\right)+\mathrm{1}\right)+\right. \\ $$$$\left(\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+..+\left(\mathrm{2}{r}−\mathrm{2}\right)+\mathrm{2}\right)+\right. \\ $$$$\left(\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+..+\left(\mathrm{2}{r}−\mathrm{2}\right)+\mathrm{3}\right)+...\right. \\ $$$$\left(\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+..+\left(\mathrm{2}{r}−\mathrm{2}\right)+\mathrm{2}{r}−\mathrm{1}\right)\right. \\ $$$$=\left(\mathrm{2}{r}−\mathrm{1}\right)\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+..+\left(\mathrm{2}{r}−\mathrm{2}\right)\right)+\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+..+\left(\mathrm{2}{r}−\mathrm{1}\right)\right) \\ $$$$=\left(\mathrm{2}{r}−\mathrm{1}\right)\frac{\left(\mathrm{2}{r}−\mathrm{2}\right)\left(\mathrm{2}{r}−\mathrm{1}\right)}{\mathrm{2}}+\frac{\left(\mathrm{2}{r}−\mathrm{1}\right)\left(\mathrm{2}{r}\right)}{\mathrm{2}} \\ $$$$=\left({r}−\mathrm{1}\right)\left(\mathrm{2}{r}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{r}−\mathrm{1}\right){r} \\ $$$$=\left(\mathrm{2}{r}−\mathrm{1}\right)\left\{\mathrm{2}{r}^{\mathrm{2}} −\mathrm{3}{r}+\mathrm{1}+{r}\right\} \\ $$$$=\left(\mathrm{2}{r}−\mathrm{1}\right)\left(\mathrm{2}{r}^{\mathrm{2}} −\mathrm{2}{r}+\mathrm{1}\right) \\ $$$$ \\ $$$${r}^{\mathrm{4}} −\left({r}−\mathrm{1}\right)^{\mathrm{4}} \\ $$$$=\left({r}^{\mathrm{2}} −\left({r}−\mathrm{1}\right)^{\mathrm{2}} \right)\left({r}^{\mathrm{2}} +\left({r}−\mathrm{1}\right)^{\mathrm{2}} \right) \\ $$$$=\left(\mathrm{2}{r}−\mathrm{1}\right)\left(\mathrm{2}{r}^{\mathrm{2}} −\mathrm{2}{r}+\mathrm{1}\right) \\ $$

Commented by nadovic last updated on 30/Jul/21

Thank Sir

$${Thank}\:{Sir} \\ $$

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