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

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

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

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

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.

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To try a problem based on this go below.

Adding rectangular areas

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

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

Note by Sanath Balaji
9 months, 2 weeks ago

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