General way to find x=1ixn,nN\sum _{ x=1 }^{ i }{ { x }^{ n } } ,\quad n\in N.

Let x=1ixn,nN=f(n)\sum _{ x=1 }^{ i }{ { x }^{ n } } ,\quad n\in N=f(n).

Using the following fact and a bit of working around we can find f(n).

That is, (1+x)α=k=0α(αk)xk (1+x)^\alpha = \sum_{k=0}^{\alpha} \binom{\alpha}{k} x^k , where (αk)=α(α1)(αk+1)k!. \binom{\alpha}{k} = \frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}.

For finding f(n) we must know f(1),f(2),......f(n-1).

The following method illustrates the way to find the sum of 4th power of natural numbers , the same can be used for finding for any nth power.


To try a problem based on this go below.

Adding rectangular areas

Find the area bounded between y=x4y = \lfloor x\rfloor ^4, the xx-axis, x=0x = 0 and x=11x=11.

Notation: \lfloor \cdot \rfloor denotes the floor function.

Note by Sanath Balaji
4 years, 5 months ago

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There's another method also known as Faulhaber's Formula.

Arkajyoti Banerjee - 3 years, 4 months ago

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