Question Number 45720 by Tawa1 last updated on 15/Oct/18
$$\underset{\boldsymbol{\mathrm{r}}\:=\:\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\boldsymbol{\sum}}}\:\:\left(\boldsymbol{\mathrm{r}}\:+\:\mathrm{1}\right)^{\mathrm{3}} \\ $$
Commented by math khazana by abdo last updated on 15/Oct/18
$$=\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:{k}^{\mathrm{3}} =\frac{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}−\mathrm{1}\:. \\ $$
Commented by Tawa1 last updated on 15/Oct/18
$$\mathrm{How}\:\mathrm{sir}\:??. \\ $$$$\mathrm{please}\:\mathrm{explain},\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by math khazana by abdo last updated on 16/Oct/18
$${changement}\:{ofi}\:{ndice}\:{r}+\mathrm{1}={k}\:{give} \\ $$$$\sum_{{r}=\mathrm{1}} ^{{n}} \left({r}+\mathrm{1}\right)^{\mathrm{3}} =\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:{k}^{\mathrm{3}} \:{but}\:{we}\:{have}\:{proved}\:{that} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}^{\mathrm{3}} =\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\:\Rightarrow\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} {k}^{\mathrm{3}} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}+\mathrm{1}} {k}^{\mathrm{3}} −\mathrm{1}\:=\frac{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}\:−\mathrm{1}\:. \\ $$
Commented by Tawa1 last updated on 16/Oct/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by maxmathsup by imad last updated on 16/Oct/18
$${you}\:{are}\:{welcome}\:{sir}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Oct/18
$${T}_{{r}} =\left({r}+\mathrm{1}\right)^{\mathrm{3}} ={r}^{\mathrm{3}} +\mathrm{3}.{r}^{\mathrm{2}} .\mathrm{1}+\mathrm{3}.{r}.\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{3}} \\ $$$${T}_{{r}} ={r}^{\mathrm{3}} +\mathrm{3}{r}^{\mathrm{2}} +\mathrm{3}{r}+\mathrm{1} \\ $$$${T}_{\mathrm{1}} =\mathrm{1}^{\mathrm{3}} +\mathrm{3}×\mathrm{1}^{\mathrm{2}} +\mathrm{3}×\mathrm{1}+\mathrm{1} \\ $$$${T}_{\mathrm{2}} =\mathrm{2}^{\mathrm{3}} +\mathrm{3}×\mathrm{2}^{\mathrm{2}} +\mathrm{3}×\mathrm{2}+\mathrm{1} \\ $$$${T}_{\mathrm{3}} =\mathrm{3}^{\mathrm{3}} +\mathrm{3}×\mathrm{3}^{\mathrm{2}} +\mathrm{3}×\mathrm{3}+\mathrm{1} \\ $$$${T}_{\mathrm{4}} =\mathrm{4}^{\mathrm{3}} +\mathrm{3}×\mathrm{4}^{\mathrm{2}} +\mathrm{3}×\mathrm{4}+\mathrm{1} \\ $$$$… \\ $$$$… \\ $$$${T}_{{n}} ={n}^{\mathrm{3}} +\mathrm{3}×{n}^{\mathrm{2}} +\mathrm{3}×{n}+\mathrm{1} \\ $$$${add}\:{them}… \\ $$$${S}_{{n}} =\left(\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +…+{n}^{\mathrm{3}} \right)+\mathrm{3}\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} …+{n}^{\mathrm{2}} \right)+ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{3}\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{n}\right)+\left(\mathrm{1}+\mathrm{1}+\mathrm{1}…{upto}\:{n}\:{times}\right) \\ $$$${S}_{{n}} =\left\{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right\}^{\mathrm{2}} +\mathrm{3}×\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}+\mathrm{3}×\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+ \\ $$$$\:\:\:{n}×\mathrm{1}\:\:{ans} \\ $$$$ \\ $$
Commented by Tawa1 last updated on 16/Oct/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by MrW3 last updated on 16/Oct/18
$${generally}\:{for}\:\mathrm{1}\leqslant{p}\leqslant{n}, \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left({r}+{p}\right)^{\mathrm{3}} =\underset{{k}={p}+\mathrm{1}} {\overset{{p}+{n}} {\sum}}{k}^{\mathrm{3}} =\underset{{k}=\mathrm{1}} {\overset{{p}+{n}} {\sum}}{k}^{\mathrm{3}} −\underset{{k}=\mathrm{1}} {\overset{{p}} {\sum}}{k}^{\mathrm{3}} \\ $$$$=\frac{\left({p}+{n}\right)^{\mathrm{2}} \left({p}+{n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}−\frac{{p}^{\mathrm{2}} \left({p}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$
Commented by Tawa1 last updated on 16/Oct/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$