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[tex]\frac{n(n+1)}{2}, \frac{n(n+1)(2n+1)}{6} ,(\frac{n(n+1)}{2})^2[/tex]

How were they evolved?

And what about [tex]i^4, i^5, i^6,...[/tex]

Thankyou everybody;;

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- Thread starter Brunll
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- #1

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[tex]\frac{n(n+1)}{2}, \frac{n(n+1)(2n+1)}{6} ,(\frac{n(n+1)}{2})^2[/tex]

How were they evolved?

And what about [tex]i^4, i^5, i^6,...[/tex]

Thankyou everybody;;

- #2

HallsofIvy

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Those are, of course, [itex]\sum_{i=1}^n i[/itex], [itex]\sum_{i=1}^n i^2[/itex], and [itex]\sum_{i=1}^n i^3[/itex], respectively. You can sum such things by using "Newton's Divided Difference" formula. It is a discrete version of Taylor's series. If f(n) is a function defined on the non-negative integers, [itex]\Delta f[/itex] is the "first difference", f(n)- f(n-1), [itex]\Delta^2 f[/itex] is the "second difference", [itex]\Delta f(n)- \Delta f(n-1)[/itex], [itex]\Delta^3 f[/itex] is the "third difference", etc. then

[tex]\frac{n(n+1)}{2}, \frac{n(n+1)(2n+1)}{6} ,(\frac{n(n+1)}{2})^2[/tex]

How were they evolved?

And what about [tex]i^4, i^5, i^6,...[/tex]

Thankyou everybody;;

[tex]f(n)= f(0)+ \Delta f(0) n+ \frac{\Delta^2 f(0)}{2!} n(n-1)+ \frac{\Delta^3 f(0)}{3!} n(n-1)(n-2)+ \cdot\cdot\cdot[/itex]

In particular, for a sum of terms in involving n

A "short cut" to finding [itex]\sum_{i=1}^n i^4[/itex] is to set it equal to An

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