How do you evaluate #5\sum _ { p = 3} ^ { 9} ( 4p ^ { 2} - 2)#?

3 Answers
Sridhar V.
Jan 18, 2018

#color(Blue)(5\sum _ { p = 3} ^ { 9} ( 4p ^ { 2} - 2)=5530#

Explanation:

Given:

#5\sum _ { p = 3} ^ { 9} ( 4p ^ { 2} - 2)#

We can expand this series as:

#rArr 5*[(4*3^2 - 2) + (4*4^2 - 2) + (4*5^2 - 2) + (4*6^2 - 2) +(4*7^2 - 2)+(4*8^2 - 2)+(4*9^2 - 2)]#

#rArr 5*[(4*9 - 2) + (4*16 - 2) + (4*25 - 2) + (4*36 - 2) +(4*49 - 2)+(4*64 - 2)+(4*81 - 2)]#

#rArr 5*[(36 - 2) + (64 - 2) + (100 - 2) + (144 - 2) +(196 - 2)+(256 - 2)+(324 - 2)]#

#rArr 5*[34 + 62 + 98 + 142 +194+254+322]#

#rArr 5*[1106]#

#rArr 5530#

Hence,

#color(Blue)(5\sum _ { p = 3} ^ { 9} ( 4p ^ { 2} - 2)=5530#

NickTheTurtle
Jan 18, 2018

#5sum_(p=3)^9(4p^2-2)=5530#

Explanation:

#\ \ \ \ \ \ 5sum_(p=3)^9(4p^2-2)#
#=5(sum_(p=3)^9(4p^2)-sum_(p=3)^9(2))#
#=5(4sum_(p=3)^9(p^2)-sum_(p=3)^9(2))#
#=5(4(sum_(p=0)^9(p^2)-sum_(p=0)^2(p^2))-sum_(p=3)^9(2))#

Note that #sum_(p=3)^9 2# basically adds #9-3+1=7# number of #2#'s together. Thus, #sum_(p=3)^9 2=2*7=14#.

Also recall that #sum_(i=0)^Ni^2=(N(2N+1)(N+1))/6#. Then, the above equates to
#=5(4((9(2*9+1)(9+1))/6-(2(2*2+1)(2+1))/6)-14)#
#=5530#

Jim G.
Jan 18, 2018

#5530#

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#

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