Find-the-formular-for-the-sum-of-the-first-kth-power-of-natural-number-

Question Number 44892 by Tawa1 last updated on 06/Oct/18

Find the formular for the sum of the first kth power of natural number

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Oct/18

d o t h e q u e s t i o n m e a n

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

p l s r e p l y

Commented by Tawa1 last updated on 06/Oct/18

Yes sir . exactly, Thanks for your help sir . God bless you

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Oct/18

Commented by Tawa1 last updated on 06/Oct/18

God bless you sir . i appreciate

Commented by Tawa1 last updated on 06/Oct/18

Sir, am trying to understand the theorem but am confused .

Can you help me to use the theorem to find the sum of n terms

1^{5} + 2^{5} + 3^{5} + 4^{5} + \dots or any example .

Please, i will read through your example and understand . God bless you

sir . And thanks for everytime

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Oct/18

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

t o f i n d S_{5} w e n e e d t h e v a l u e o f S_{4}

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

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

S_{4} = 4 \int \frac{n^{4} + 2 n^{3} + n^{2}}{4} d n + {n B}_{4}

= \frac{n^{5}}{5} + \frac{2 n^{4}}{4} + \frac{n^{3}}{3} + n \times \frac{- 1}{30}

= \frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}

S_{5} = 5 \int S_{4} d n + {n B}_{5}

S_{5} = 5 \int \frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30} d n + {n B}_{5}

S_{5} = 5 {\frac{n^{6}}{30} + \frac{n^{5}}{10} + \frac{n^{4}}{12} - \frac{n^{2}}{60}} + {n B}_{5}

S_{5} = \frac{n^{6}}{6} + \frac{n^{5}}{2} + \frac{5 n^{4}}{12} - \frac{n^{2}}{12} + {n B}_{5}

v a l u e o f B_{5} = 0 s o

1^{5} + 2^{5} + 3^{5} + \dots + n^{5} = \frac{n^{6}}{6} + \frac{n^{5}}{2} + \frac{5 n^{4}}{12} - \frac{n^{2}}{12}

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Oct/18

Commented by Tawa1 last updated on 06/Oct/18

Wow, God bless you sir . Thanks for your time

Commented by Tawa1 last updated on 06/Oct/18

I really understand now . God bless you sir

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