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.

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.

#Calculus

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

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

Arkajyoti Banerjee - 3 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

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