prove-that1-3-2-3-3-3-n-3-n-2-n-1-2-4-for-every-natural-number-n-

Question Number 45585 by pieroo last updated on 14/Oct/18

prove {that 1}^{3} + 2^{3} + 3^{3} + \dots + n^{3} = \frac{n^{2} {(n + 1)}^{2}}{4} for every

natural number n

Commented by maxmathsup by imad last updated on 14/Oct/18

l e t p (x) = \frac{x^{2} {(x - 1)}^{2}}{4} w e h a v e p (x + 1) - p (x) = \frac{{(x + 1)}^{2} x^{2}}{4} - \frac{x^{2} {(x - 1)}^{2}}{4}

= \frac{x^{2}}{4} {{(x + 1)}^{2} - {(x - 1)}^{2}} = \frac{x^{2}}{4} {x^{2} + 2 x + 1 - x^{2} + 2 x - 1} = x^{3} \Rightarrow

\sum_{k = 1}^{n} (p (k + 1) - p (k)) = \sum_{k = 1}^{n} k^{3} \Rightarrow \sum_{k = 1}^{n} k^{3} = p (n + 1) - p (1

= \frac{{(n + 1)}^{2} n^{2}}{4} = \frac{n^{2} {(n + 1)}^{2}}{4} .

Commented by pieroo last updated on 14/Oct/18

thanks very much sir .

Commented by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18

a n o t h e r g e n e r a l m e t h o d \dots

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

f o r m u l s i s S_{r} = r \int S_{r - 1} d n + {n B}_{r}

n o w i f w e k n o w s_{1} = 1 + 2 + 3 + \dots + n

w e c a n f i n d S_{2} = 1^{2} + 2^{2} + 3^{2} + \dots + n^{2}

S_{3} = 1^{3} + 2^{3} + 3^{3} + . . + n^{3}

S_{40} = 1^{40} + 2^{40} + 3^{40} + \dots + n^{40}

n o w s_{1} = \frac{n (n + 1)}{2} = \frac{n^{2}}{2} + \frac{n}{2}

S_{r} = r \int S_{r - 1} d n + {n B}_{r}

s o S_{2} = 2 \int S_{1} d n + {n B}_{2}

= 2 \int (\frac{n^{2}}{2} + \frac{n}{2}) d n + {n B}_{2}

= 2 [\frac{n^{3}}{6} + \frac{n^{2}}{4}] + n \times \frac{1}{6} B_{2} = \frac{1}{6} (s e e t a b l e)

= \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6} = \frac{2 n^{3} + 3 n^{2} + n}{6}

= \frac{n (2 n^{2} + 3 n + 1)}{6} = \frac{n}{6} (2 n^{2} + 2 n + n + 1)

= \frac{n (n + 1) (2 n + 1)}{6}

u s i n g t h i s w e c a n f i n d S_{3} = 3 \int S_{2} d n + {n B}_{3}

S_{3} = 3 \int \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6} + n \times 0 B_{3} = 0

= 3 (\frac{n^{4}}{12} + \frac{n^{3}}{6} + \frac{n^{2}}{12}) = \frac{n^{4}}{4} + \frac{n^{3}}{2} + \frac{n^{2}}{4} = \frac{n^{2}}{4} (n^{2} + 2 n + 1)

= \frac{n^{2}}{4} {(n + 1)}^{2} = {\frac{n (n + 1)}{2}}^{2} i s t h e a n s w e r

Commented by pieroo last updated on 14/Oct/18

wow, i love this

Commented by pieroo last updated on 14/Oct/18

Sir thank you very much, but please help me with

Question number 45353

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