Summation

Algebra Level 4

If ∑ r = 1 n 6 r 3 + 18 r 2 + 12 r + 2 r ( r + 1 ) ( r + 2 ) = a n 3 + b n 2 + c n 2 ( n + d ) ( n + e ) \displaystyle\sum_{r=1}^n\dfrac{6r^3+18r^2+12r+2}{r(r+1)(r+2)}= \dfrac{an^3 +bn^2+cn}{2(n+d)(n+e)}

for positive integers a , b , c , d a,b,c,d and e e , such that e > d e >d

then evaluate 2 e ( a + b ) c + d \dfrac{2e(a+b)}{c+d}


The answer is 7.

Nitish Joshi by Nitish Joshi , 22, India

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1 solution

Rishabh Jain
Jan 1, 2016

∑ r = 1 n \displaystyle \sum_{r=1}^n 6 ( r 3 + 3 r 2 + 2 r ) + 2 r 3 + 3 r 2 + 2 r \frac {6(r^3+3r^2+2r)+2}{r^3+3r^2+2r}

= ∑ r = 1 n 6 + 2 r ( r + 1 ) ( r + 2 ) =\displaystyle \sum_{r=1}^n 6+ \frac{2}{r(r+1)(r+2)} = 6 n + 1 2 − 1 ( n + 1 ) ( n + 2 ) =6n+\frac {1}{2} -\frac{1}{(n+1)(n+2)}

= 12 n 3 + 37 n 2 + 27 n 2 ( n + 1 ) ( n + 2 ) =\frac{12n^3+37n^2+27n}{2(n+1)(n+2)}

a=12,b=37,c=27,d=1,e=2.

Therefore 2 e ( a + b ) c + d = 2 • 2 ( 12 + 37 ) 27 + 1 = 7 \frac {2e(a+b)}{c+d}= \frac {2•2(12+37)}{27+1}= 7

3 Helpful 0 Interesting 0 Brilliant 0 Confused

I think we need some extra condition since d=2,e=1 is also a possible case.

bhavay kukreja - 5 years, 1 month ago

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Exactly... Some condition need to be specified....

Rishabh Jain - 5 years, 1 month ago

I have edited the problem accordingly. Thanks a lot for pointing it out!

Nitish Joshi - 5 years, 1 month ago

I did the same

Aditya Kumar - 5 years, 1 month ago

It was stated that e > d in the antecedent

Maddie Dog - 2 years, 4 months ago

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