# how to calculate the sum of n natural numbers of powers 4,5,6,7.?

like 1^4+2^4+...n^4 and similarly for other powers

### 1 Answer

- MadhukarLv 79 years agoFavourite answer
Knowing

1 + 2 + 3 + 4 + ... + n = (1/2)n(n+1)

1^2 + 2^2 + 3^2 + 4^2 + ... + n^2 = (1/6) n(n+1)(2n+1) and

1^3 + 2^3 + 3^3 + 4^3 + ... +n^3 = (1/4)n^2 (n+1)^2,

the sum of 4th powers of natural numbers can be found as under.

n^5 - (n-1)^5 = n^5 - (n^5 - 5n^4 + 10n^3 - 10n^2 + 5n - 1)

=> n^5 - (n-1)^5 = 5n^4 - 10n^3 + 10n^2 - 5n + 1

Plugging n = 1, 2, 3, 4, ... n

1^5 - 0^5 = 5*(1^4) - 10*(1^3) + 10*(1^2) - 5*(1) + 1

2^5 - 1^5 = 5*(2^4) - 10*(2^3) + 10*(2^2)- 5*(2) + 1

3^5 - 2^5 = 5*(3^4) - 10*(3^3) + 10*(3^2) - 5*(3) + 1

4^5 - 3^5 = 5*(4^4) - 10*(4^3) + 10*(4^2) - 5*(4) + 1

.......

.....

n^5 - (n-1)^5 = 5(n^4) - 10*(n^3) + 10*(n^2) - 5*(n) + 1

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Adding all,

n^5 = 5 Σn^4 - 10 Σn^3 + 10 Σn^2 - 5 Σn + Σ 1

=> 5 Σn^4

= n^5 + 10 Σn^3 - 10 Σn^2 + 5 Σn - n

= n^5 + (10/4)n^2 (n+1)^2 - (10/6) n (n+1) (2n+1) + (5/2) n (n+1) - n

= (1/12) [12n^5 + 30n^4 + 60n^3 + 30n^2 - 20(2n^3 + 3n^2 + n) + 30 (n^2 + n) - 12n]

= (1/12) (12n^5 + 30n^4 + 20n^3 - 2n)

= (1/6) (6n^5 + 15n^4 + 10n^3 - n)

=> Σn^4 = (1/30) (6n^5 + 15n^4 + 10n^3 - 30n^2 - 31n).

Similarly, to find Σ n^5, start by writing n^6 - (n-1)^6 = .... and so on.

your approach is great but the formula is apparently wrong. Put any value of n and you'll find that it gives a sum less by a figure of n(n+1).

Maybe you missed something in the formula...

Adding n(n+1) to it gives the actual formula to be:

(1/30)[6n^5+15n^4+10n^3-n]