r-1-n-r-1-3-

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} + 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

\dots

\dots

T_{n} = n^{3} + 3 \times n^{2} + 3 \times n + 1

a d d t h e m \dots

S_{n} = (1^{3} + 2^{3} + 3^{3} + \dots + n^{3}) + 3 (1^{2} + 2^{2} + 3^{2} \dots + n^{2}) +

3 (1 + 2 + 3 + \dots + n) + (1 + 1 + 1 \dots 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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