If n is a positive integers,show that (n+1)^2+(n+2)^2+....+4n^2=n/6(2n+1)(7n+1)?

Question

3226 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-25T11:57:42

See the explanation below

We need

#sum_(n=1)^n n^2=n/6(n+1)(2n+1)#

Therefore,

#sum_(n=1)^(2n)n^2 = ( 2n)/(6)(2n+1)(4n+1)#

#sum_(n+1)^(2n) n^2 = sum_(n=1)^(2n)n^2 - sum_(n=1)^n n^2#

#=( 2n)/(6)(2n+1)(4n+1) - n/6(n+1)(2n+1)#

#=n/6(2(2n+1)(4n+1)- (n+1)(2n+1))#

#=n/6((16n^2+12n+2)-(2n^2+3n+1))#

#=n/6(14n^2+9n+1)#

#=n/6(7n+1)(2n+1)#