𝚺_(r = 1) ^n (r + 1)^3


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

$\underset{r = 1}{\sum^{n}} {(r + 1)}^{3}$
Commented by math khazana by abdo last updated on 15/Oct/18

$= \sum_{k = 2}^{n + 1} k^{3} = \frac{{(n + 1)}^{2} {(n + 2)}^{2}}{4} - 1 .$
Commented by Tawa1 last updated on 15/Oct/18

$How sir ? ? .$ $please explain, God bless you sir$
Commented by math khazana by abdo last updated on 16/Oct/18

$c h a n g e m e n t o f i n d i c e r + 1 = k g i v e$ $\sum_{r = 1}^{n} {(r + 1)}^{3} = \sum_{k = 2}^{n + 1} k^{3} b u t w e h a v e p r o v e d t h a t$ $\sum_{k = 1}^{n} k^{3} = \frac{n^{2} {(n + 1)}^{2}}{4} \Rightarrow \sum_{k = 2}^{n + 1} k^{3}$ $= \sum_{k = 1}^{n + 1} k^{3} - 1 = \frac{{(n + 1)}^{2} {(n + 2)}^{2}}{4} - 1 .$
Commented by Tawa1 last updated on 16/Oct/18

$God bless you sir$
Commented by maxmathsup by imad last updated on 16/Oct/18

$y o u a r e w e l c o m e s i r .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 16/Oct/18
	$T_r =(r+1)^3 =r^3 +3.r^2 .1+3.r.1^2 +1^3 T_r =r^3 +3r^2 +3r+1 T_1 =1^3 +3×1^2 +3×1+1 T_2 =2^3 +3×2^2 +3×2+1 T_3 =3^3 +3×3^2 +3×3+1 T_4 =4^3 +3×4^2 +3×4+1 ... ... T_n =n^3 +3×n^2 +3×n+1 add them... S_n =(1^3 +2^3 +3^3 +...+n^3 )+3(1^2 +2^2 +3^2 ...+n^2 )+ 3(1+2+3+...+n)+(1+1+1...upto n times) S_n ={((n(n+1))/2)}^2 +3×((n(n+1)(2n+1))/6)+3×((n(n+1))/2)+ n×1 ans$
	$T_{r} = {(r + 1)}^{3} = r^{3} + 3 . r^{2} . 1 + 3 . r . 1^{2} + 1^{3}$ $T_{r} = r^{3} + 3 r^{2} + 3 r + 1$ $T_{1} = 1^{3} + 3 \times 1^{2} + 3 \times 1 + 1$ $T_{2} = 2^{3} + 3 \times 2^{2} + 3 \times 2 + 1$ $T_{3} = 3^{3} + 3 \times 3^{2} + 3 \times 3 + 1$ $T_{4} = 4^{3} + 3 \times 4^{2} + 3 \times 4 + 1$ $. . .$ $. . .$ $T_{n} = n^{3} + 3 \times n^{2} + 3 \times n + 1$ $a d d t h e m . . .$ $S_{n} = (1^{3} + 2^{3} + 3^{3} + . . . + n^{3}) + 3 (1^{2} + 2^{2} + 3^{2} . . . + n^{2}) +$ $3 (1 + 2 + 3 + . . . + n) + (1 + 1 + 1 . . . u p t o n t i m e s)$ $S_{n} = {\frac{n (n + 1)}{2}}^{2} + 3 \times \frac{n (n + 1) (2 n + 1)}{6} + 3 \times \frac{n (n + 1)}{2} +$ $n \times 1 a n s$
	Commented by Tawa1 last updated on 16/Oct/18

	$God bless you sir$
	Answered by MrW3 last updated on 16/Oct/18

	$g e n e r a l l y f o r 1 ⩽ p ⩽ n,$ $\underset{r = 1}{\sum^{n}} {(r + p)}^{3} = \underset{k = p + 1}{\sum^{p + n}} k^{3} = \underset{k = 1}{\sum^{p + n}} k^{3} - \underset{k = 1}{\sum^{p}} k^{3}$ $= \frac{{(p + n)}^{2} {(p + n + 1)}^{2}}{4} - \frac{p^{2} {(p + 1)}^{2}}{4}$
	Commented by Tawa1 last updated on 16/Oct/18

	$God bless you sir$
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