Prove that $n \sum r = 1 r^{5} = \frac{1}{12} n^{2} {(n + 1)}^{2} (2 n^{2} + 2 n - 1)$ $sum_(r=1)^nr^5=1/12n^2(n+1)^2(2n^2+2n-1)$ using binomial theorem?

Question

Prove that $n \sum r = 1 r^{5} = \frac{1}{12} n^{2} {(n + 1)}^{2} (2 n^{2} + 2 n - 1)$ $sum_(r=1)^nr^5=1/12n^2(n+1)^2(2n^2+2n-1)$ using binomial theorem?

I'm confident of the method up till the step
${(n + 1)}^{6} = 6 n \sum r = 1 r^{5} + 15 n \sum r = 1 r^{4} + 20 n \sum r = 1 r^{3} + 15 n \sum r = 1 r^{2} + 6 n \sum r = 1 r + n + 1$ $(n+1)^6=6sum_(r=1)^nr^5+15sum_(r=1)^nr^4+20sum_(r=1)^nr^3+15sum_(r=1)^nr^2+6sum_(r=1)^nr+n+1$
but I'm not sure of my algebra afterwards;; I keep reaching different answers. Thanks!

1 Answer

Impact of this question

2370 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-06T17:33:33

See below.

Explanation:

By brute force

Making

$f (n) = c_{0} + c_{1} n + c_{2} n^{2} + c_{3} n^{3} + c_{4} n^{4} + c_{5} n^{5} + c_{6} n^{6} - (n \sum k = 1 k^{5}) = 0$ $f(n) =c_0+c_1 n+c_2 n^2+c_3 n^3+c_4 n^4+c_5 n^5+c_6 n^6 -(sum_(k=1)^n k^5)= 0$

This should be an identity so

${ (f(1)=c_0 + c_1 + c_2 + c_3 + c_4 + c_5 + c_6 =1), (f(2)=c_0 + 2 c_1 + 4 c_2 + 8 c_3 + 16 c_4 + 32 c_5 + 64 c_6 =33), (f(3)=c_0 + 3 c_1 + 9 c_2 + 27 c_3 + 81 c_4 + 243 c_5 + 729 c_6 = 276), ( f(4)=c_0 + 4 c_1 + 16 c_2 + 64 c_3 + 256 c_4 + 1024 c_5 + 4096 c_6 = 1300 ), ( f(5)=c_0 + 5 c_1 + 25 c_2 + 125 c_3 + 625 c_4 + 3125 c_5 + 15625 c_6 = 4425), (f(6)=c_0 + 6 c_1 + 36 c_2 + 216 c_3 + 1296 c_4 + 7776 c_5 + 46656 c_6 = 12201), (f(7)=c_0 + 7 c_1 + 49 c_2 + 343 c_3 + 2401 c_4 + 16807 c_5 + 117649 c_6 = 29008):}$

Solving this linear system we get

$c_0=0,c_1=0,c_2=-1/12,c_3=0,c_4=5/12,c_5=1/2, c_6 = 1/6$ or

$sum_(k=1)^n k^5=-n^2/12 + (5 n^4)/12 + n^5/2 + n^6/6$ or

$sum_(k=1)^n k^5=1/12 n^2 (n+1)^2 (2 n^2+2n-1)$

Science

Math

Humanities

... and beyond

Prove that $n \sum r = 1 r^{5} = \frac{1}{12} n^{2} {(n + 1)}^{2} (2 n^{2} + 2 n - 1)$ $sum_(r=1)^nr^5=1/12n^2(n+1)^2(2n^2+2n-1)$ using binomial theorem?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Prove that sum_(r=1)^nr^5=1/12n^2(n+1)^2(2n^2+2n-1)n∑r=1r5=112n2(n+1)2(2n2+2n−1)sum_(r=1)^nr^5=1/12n^2(n+1)^2(2n^2+2n-1) using binomial theorem?

1 Answer

Explanation:

Related questions

Impact of this question

Prove that $n \sum r = 1 r^{5} = \frac{1}{12} n^{2} {(n + 1)}^{2} (2 n^{2} + 2 n - 1)$ $sum_(r=1)^nr^5=1/12n^2(n+1)^2(2n^2+2n-1)$ using binomial theorem?