Menu Close

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

Tinku Tara 0 Comments
Pin




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

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *