Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 45720 by Tawa1 last updated on 15/Oct/18

𝚺_(r = 1) ^n (r + 1)^3

$$\underset{\boldsymbol{\mathrm{r}}\:=\:\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\boldsymbol{\sum}}}\:\:\left(\boldsymbol{\mathrm{r}}\:+\:\mathrm{1}\right)^{\mathrm{3}} \\ $$

Commented by math khazana by abdo last updated on 15/Oct/18

=Σ_(k=2) ^(n+1) k^3 =(((n+1)^2 (n+2)^2 )/4)−1 .

$$=\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:{k}^{\mathrm{3}} =\frac{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}−\mathrm{1}\:. \\ $$

Commented by Tawa1 last updated on 15/Oct/18

How sir ??. please explain, God bless you sir

$$\mathrm{How}\:\mathrm{sir}\:??. \\ $$$$\mathrm{please}\:\mathrm{explain},\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by math khazana by abdo last updated on 16/Oct/18

changement ofi ndice r+1=k give Σ_(r=1) ^n (r+1)^3 =Σ_(k=2) ^(n+1) k^3 but we have proved that Σ_(k=1) ^n k^3 =((n^2 (n+1)^2 )/4) ⇒Σ_(k=2) ^(n+1) k^3 =Σ_(k=1) ^(n+1) k^3 −1 =(((n+1)^2 (n+2)^2 )/4) −1 .

$${changement}\:{ofi}\:{ndice}\:{r}+\mathrm{1}={k}\:{give} \\ $$$$\sum_{{r}=\mathrm{1}} ^{{n}} \left({r}+\mathrm{1}\right)^{\mathrm{3}} =\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:{k}^{\mathrm{3}} \:{but}\:{we}\:{have}\:{proved}\:{that} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}^{\mathrm{3}} =\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\:\Rightarrow\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} {k}^{\mathrm{3}} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}+\mathrm{1}} {k}^{\mathrm{3}} −\mathrm{1}\:=\frac{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}\:−\mathrm{1}\:. \\ $$

Commented by Tawa1 last updated on 16/Oct/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by maxmathsup by imad last updated on 16/Oct/18

you are welcome sir.

$${you}\:{are}\:{welcome}\:{sir}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Oct/18

T_r =(r+1)^3 =r^3 +3.r^2 .1+3.r.1^2 +1^3 T_r =r^3 +3r^2 +3r+1 T_1 =1^3 +3×1^2 +3×1+1 T_2 =2^3 +3×2^2 +3×2+1 T_3 =3^3 +3×3^2 +3×3+1 T_4 =4^3 +3×4^2 +3×4+1 ... ... T_n =n^3 +3×n^2 +3×n+1 add them... S_n =(1^3 +2^3 +3^3 +...+n^3 )+3(1^2 +2^2 +3^2 ...+n^2 )+ 3(1+2+3+...+n)+(1+1+1...upto n times) S_n ={((n(n+1))/2)}^2 +3×((n(n+1)(2n+1))/6)+3×((n(n+1))/2)+ n×1 ans

$${T}_{{r}} =\left({r}+\mathrm{1}\right)^{\mathrm{3}} ={r}^{\mathrm{3}} +\mathrm{3}.{r}^{\mathrm{2}} .\mathrm{1}+\mathrm{3}.{r}.\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{3}} \\ $$$${T}_{{r}} ={r}^{\mathrm{3}} +\mathrm{3}{r}^{\mathrm{2}} +\mathrm{3}{r}+\mathrm{1} \\ $$$${T}_{\mathrm{1}} =\mathrm{1}^{\mathrm{3}} +\mathrm{3}×\mathrm{1}^{\mathrm{2}} +\mathrm{3}×\mathrm{1}+\mathrm{1} \\ $$$${T}_{\mathrm{2}} =\mathrm{2}^{\mathrm{3}} +\mathrm{3}×\mathrm{2}^{\mathrm{2}} +\mathrm{3}×\mathrm{2}+\mathrm{1} \\ $$$${T}_{\mathrm{3}} =\mathrm{3}^{\mathrm{3}} +\mathrm{3}×\mathrm{3}^{\mathrm{2}} +\mathrm{3}×\mathrm{3}+\mathrm{1} \\ $$$${T}_{\mathrm{4}} =\mathrm{4}^{\mathrm{3}} +\mathrm{3}×\mathrm{4}^{\mathrm{2}} +\mathrm{3}×\mathrm{4}+\mathrm{1} \\ $$$$... \\ $$$$... \\ $$$${T}_{{n}} ={n}^{\mathrm{3}} +\mathrm{3}×{n}^{\mathrm{2}} +\mathrm{3}×{n}+\mathrm{1} \\ $$$${add}\:{them}... \\ $$$${S}_{{n}} =\left(\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +...+{n}^{\mathrm{3}} \right)+\mathrm{3}\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} ...+{n}^{\mathrm{2}} \right)+ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{3}\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+...+{n}\right)+\left(\mathrm{1}+\mathrm{1}+\mathrm{1}...{upto}\:{n}\:{times}\right) \\ $$$${S}_{{n}} =\left\{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right\}^{\mathrm{2}} +\mathrm{3}×\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}+\mathrm{3}×\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+ \\ $$$$\:\:\:{n}×\mathrm{1}\:\:{ans} \\ $$$$ \\ $$

Commented by Tawa1 last updated on 16/Oct/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by MrW3 last updated on 16/Oct/18

generally for 1≤p≤n, Σ_(r=1) ^n (r+p)^3 =Σ_(k=p+1) ^(p+n) k^3 =Σ_(k=1) ^(p+n) k^3 −Σ_(k=1) ^p k^3 =(((p+n)^2 (p+n+1)^2 )/4)−((p^2 (p+1)^2 )/4)

$${generally}\:{for}\:\mathrm{1}\leqslant{p}\leqslant{n}, \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left({r}+{p}\right)^{\mathrm{3}} =\underset{{k}={p}+\mathrm{1}} {\overset{{p}+{n}} {\sum}}{k}^{\mathrm{3}} =\underset{{k}=\mathrm{1}} {\overset{{p}+{n}} {\sum}}{k}^{\mathrm{3}} −\underset{{k}=\mathrm{1}} {\overset{{p}} {\sum}}{k}^{\mathrm{3}} \\ $$$$=\frac{\left({p}+{n}\right)^{\mathrm{2}} \left({p}+{n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}−\frac{{p}^{\mathrm{2}} \left({p}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$

Commented by Tawa1 last updated on 16/Oct/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com