Question #ff053

Question

2249 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-01T09:28:41

#1/3 n ( 4 n^2-1)#

We know that

#S_(2n)=sum_(k=1)^n(2k)^2+sum_(k=0)^(n-1)(2k+1)^2=sum_(k=1)^(2n)k^2#

then

#sum_(k=0)^(n-1)(2k+1)^2=S_(2n)-4 S_n#

Here

#S_n=n^3/3+n^2/2+n/6#

so

#S_(2n)-4 S_n=1/3 n ( 4 n^2-1)#