How do you evaluate 5\sum _ { p = 3} ^ { 9} ( 4p ^ { 2} - 2)59∑p=3(4p2−2)?

3 Answers
Sridhar V.
Jan 18, 2018

color(Blue)(5\sum _ { p = 3} ^ { 9} ( 4p ^ { 2} - 2)=553059∑p=3(4p2−2)=5530

Explanation:

Given:

5\sum _ { p = 3} ^ { 9} ( 4p ^ { 2} - 2)59∑p=3(4p2−2)

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)]⇒5⋅[(4⋅32−2)+(4⋅42−2)+(4⋅52−2)+(4⋅62−2)+(4⋅72−2)+(4⋅82−2)+(4⋅92−2)]

rArr 5*[(4*9 - 2) + (4*16 - 2) + (4*25 - 2) + (4*36 - 2) +(4*49 - 2)+(4*64 - 2)+(4*81 - 2)]⇒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)]⇒5⋅[(36−2)+(64−2)+(100−2)+(144−2)+(196−2)+(256−2)+(324−2)]

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

rArr 5*[1106]⇒5⋅[1106]

rArr 5530⇒5530

Hence,

color(Blue)(5\sum _ { p = 3} ^ { 9} ( 4p ^ { 2} - 2)=553059∑p=3(4p2−2)=5530

NickTheTurtle
Jan 18, 2018

5sum_(p=3)^9(4p^2-2)=553059∑p=3(4p2−2)=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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