Question Number 65199 by rajesh4661kumar@gmail.com last updated on 26/Jul/19
Answered by Tanmay chaudhury last updated on 26/Jul/19
![(1),(2,3,4),(5,6,7,8,9),(10,11,12,13,14,15,16)... 1st group contains→ 1 terms 2nd group contains → 3 terms 3rd group contains→ 5terms 4th grouo contains → 7terms ..... ..... nth group contains →{1+(n−1)2}→ 2n−1 terms determination of first term of nth group let t_n =be the first term of nth group taking the first term of each group S=1+2+5+10+17+....+t_n (←total nterms) S= 1+2 +5+10+....+t_(n−1) +t_n (←total n terms) substructing 0=1+1+3+5+7+....−t_n (←total n+1 terms) t_n =1+1+3+5+7.....(←total nterms) t_n =1+(1+3+5+7+...←total n−1 terms) t_n =1+((n−1)/2)[2×1+(n−1−1)×2] t_n =1+((n−1)/2)×(2n−2) t_n =n^2 −2n+2 so sum of terms in nth group is S_(nth) =((2n−1)/2)[2t_n +(2n−1−1)×1] =((2n−1)/2)[2n^2 −4n+4+2n−2] =(2n−1)(n^2 −n+1) =2n^3 −2n^2 +2n−n^2 +n−1 =n^3 +n^3 −3n^2 +3n−1 =(n)^3 +(n−1)^3 proved](https://www.tinkutara.com/question/Q65208.png)
$$\left(\mathrm{1}\right),\left(\mathrm{2},\mathrm{3},\mathrm{4}\right),\left(\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right),\left(\mathrm{10},\mathrm{11},\mathrm{12},\mathrm{13},\mathrm{14},\mathrm{15},\mathrm{16}\right)… \\ $$$$\mathrm{1}{st}\:{group}\:\:{contains}\rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:{terms} \\ $$$$\mathrm{2}{nd}\:{group}\:{contains}\:\rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\:{terms} \\ $$$$\mathrm{3}{rd}\:\:\:{group}\:\:\:{contains}\rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}{terms} \\ $$$$\mathrm{4}{th}\:\:\:\:{grouo}\:{contains}\:\:\:\rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{7}{terms} \\ $$$$….. \\ $$$$….. \\ $$$${nth}\:{group}\:{contains}\:\:\:\:\rightarrow\left\{\mathrm{1}+\left({n}−\mathrm{1}\right)\mathrm{2}\right\}\rightarrow\:\:\:\:\:\mathrm{2}{n}−\mathrm{1}\:{terms} \\ $$$${determination}\:{of}\:{first}\:{term}\:{of}\:{nth}\:{group} \\ $$$${let}\:{t}_{{n}} ={be}\:{the}\:{first}\:{term}\:{of}\:{nth}\:{group} \\ $$$${taking}\:{the}\:{first}\:{term}\:{of}\:{each}\:{group} \\ $$$${S}=\mathrm{1}+\mathrm{2}+\mathrm{5}+\mathrm{10}+\mathrm{17}+….+{t}_{{n}} \:\:\:\:\:\left(\leftarrow{total}\:{nterms}\right) \\ $$$${S}=\:\:\:\:\:\:\:\:\mathrm{1}+\mathrm{2}\:+\mathrm{5}+\mathrm{10}+….+{t}_{{n}−\mathrm{1}} +{t}_{{n}} \left(\leftarrow{total}\:{n}\:{terms}\right) \\ $$$${substructing}\: \\ $$$$\mathrm{0}=\mathrm{1}+\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}+….−{t}_{{n}} \:\:\left(\leftarrow{total}\:{n}+\mathrm{1}\:{terms}\right) \\ $$$${t}_{{n}} =\mathrm{1}+\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}…..\left(\leftarrow{total}\:{nterms}\right) \\ $$$${t}_{{n}} =\mathrm{1}+\left(\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}+…\leftarrow{total}\:{n}−\mathrm{1}\:{terms}\right) \\ $$$${t}_{{n}} =\mathrm{1}+\frac{{n}−\mathrm{1}}{\mathrm{2}}\left[\mathrm{2}×\mathrm{1}+\left({n}−\mathrm{1}−\mathrm{1}\right)×\mathrm{2}\right] \\ $$$${t}_{{n}} =\mathrm{1}+\frac{{n}−\mathrm{1}}{\mathrm{2}}×\left(\mathrm{2}{n}−\mathrm{2}\right) \\ $$$${t}_{{n}} ={n}^{\mathrm{2}} −\mathrm{2}{n}+\mathrm{2} \\ $$$${so}\:{sum}\:{of}\:{terms}\:{in}\:{nth}\:{group}\:{is} \\ $$$${S}_{{nth}} =\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\left[\mathrm{2}{t}_{{n}} +\left(\mathrm{2}{n}−\mathrm{1}−\mathrm{1}\right)×\mathrm{1}\right] \\ $$$$=\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\left[\mathrm{2}{n}^{\mathrm{2}} −\mathrm{4}{n}+\mathrm{4}+\mathrm{2}{n}−\mathrm{2}\right] \\ $$$$=\left(\mathrm{2}{n}−\mathrm{1}\right)\left({n}^{\mathrm{2}} −{n}+\mathrm{1}\right) \\ $$$$=\mathrm{2}{n}^{\mathrm{3}} −\mathrm{2}{n}^{\mathrm{2}} +\mathrm{2}{n}−{n}^{\mathrm{2}} +{n}−\mathrm{1} \\ $$$$={n}^{\mathrm{3}} +{n}^{\mathrm{3}} −\mathrm{3}{n}^{\mathrm{2}} +\mathrm{3}{n}−\mathrm{1} \\ $$$$=\left({n}\right)^{\mathrm{3}} +\left({n}−\mathrm{1}\right)^{\mathrm{3}} \:\:\:{proved} \\ $$
Commented by peter frank last updated on 26/Jul/19
$${thank}\:{you} \\ $$