How do you evaluate #5\sum _ { p = 3} ^ { 9} ( 4p ^ { 2} - 2)#?
3 Answers
Explanation:
Given:
We can expand this series as:
Hence,
Explanation:
Note that
Also recall that
Explanation:
#"using the fundamental 'building blocks' for"sum#
#•color(white)(x)sum_(r=1)^n1=n#
#•color(white)(x)sum_(r=1)^nr^2=1/6n(n+1)(2n+1)#
#rArr5sum_(r=1)^n(4p^2-2)#
#=5[4sum_(r=1)^np^2-2sum_(r=1)^n1]#
#=20sum_(r=1)^np^2-10n#
#=10/3n(2n^2+3n+1)-10n#
#=20/3n^3+10n^2+10/3n-10n#
#=20/3n(n^2+3/2n-1)#
#sum_(p=3)^9=sum_(p=1)^9-sum_(p=1)^2#
#color(white)(xxx)=60(81+27/2-1)-40/3(4+3-1)#
#color(white)(xxx)=5610-80#
#color(white)(xxx)=5530#