What is the formula to find sum of cube of 1>odd natural numbers 2> even natural numbers?
Greetings, sum of cubes natural numbers from 1 to n is (n(n+1)/2)^2 ...1) sum of cubes of n even natural numbers is 2^3 + 4^3 + ... + (2n)^3 = 8(1^3 + 2^3 + ... + n^3) = 8(n(n+1)/2)^2 = 2(n(n+1))^2 ...2) sum of cubes of n odd natural numbers is the difference of 1^3 + 3^3 + ... + (2n - 1)^3 = 1^3 + 2^3 + ... + (2n)^2 - ( 2^3 + 4^3 + ... + (2n)^2) = ((2n)(2n + 1)/2)^2 - 2(n(n + 1))^2 = n^2(2n^2 - 1) Regards
Is there a formula for the sum of any power? How can we derive a formula for the sum [math]\sum_{k=1}^n k^{20}[/math]?
I don't know the exact answer about your question. But I can give you some idea about how to solve this problem.Let's say [math]S_{m,n}=\sum_{k=1}^{n}k^m[/math], e.g. [math]S_{1,n}=\sum_{k=1}^{n}k=\frac{(n+1)n}{2}[/math].If you want to get [math]S_{2,n}=\sum_{k=1}^{n}k^2[/math]. We take [math]S_{3,n}=1+\sum_{k=2}^{n}k^3[/math][math]=1+\sum_{k=1}^{n-1}(k+1)^3[/math][math]=S_{3,n}-n^3+3S_{2,n}-3n^2+3S_{1,n}-2n[/math]Cancel [math]S_{3,n}[/math] from each side, then you can solve [math]S_{2,n}=\frac{n(n+1)(2n+1)}{6}[/math].So, if you want [math]S_{20,n}=\sum_{k=1}^{n}k^{20}[/math], you do [math]S_{21,n}[/math], and after you solved [math]S_{1\dots 19,n}[/math], you'll find your answer.
How do I find the formula for 1/1*2 + 1/2*3 + ... + 1/n(n+1)?
Actually this is a telescoping series. See http://en.wikipedia.org/wiki/Telescoping... We can rewrite 1/[n(n+1) as (n+1 -n)/[n(n+1)] = 1/n - 1/(n+1) by dividing through the denominator. We therefore have 1/1 - 1/(1+1) + (1/2 - 1/(2+1) + (1/3 - 1/(3+1) + ... + (1/n - 1/(n+1). It is clear that the sum is actually 1/1 - 1/(n+1) = n/(n+1) by regrouping each term. The limit of the sum as n approaches infinity is 1.