# if m>1, then sum of the nth power of n even number is

Arun Kumar IIT Delhi
8 years ago
Hello Student,
You do this by Faulhaber's formula.
But i don’t think this is in JEE syllabus.
They will ask maximum like 3.
$\\\sum_{1}^{n} (2r)^n=??? \\=>2^n\sum_{1}^{n} (r)^n \\Now \\(n+1)^{k+1}-1=\sum_{1}^{n}(m+1)^{k+1}-(m)^{k+1} \\=_{p}^{k+1}\textrm{C}\sum_{1}^{n}r^p$
Thanks & Regards
Arun Kumar
Btech, IIT Delhi