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Worked solution

Deduce an infinite tail from a finite identity

Given ∑r=1n7r+10r(r+1)(r+2)=6−(5n+1+2n+2)\sum_{r=1}^n\dfrac{7r+10}{r(r+1)(r+2)}=6-\left(\dfrac5{n+1}+\dfrac2{n+2}\right), deduce the exact value of ∑r=22∞21r+30r(r+1)(r+2)\sum_{r=22}^\infty\dfrac{21r+30}{r(r+1)(r+2)}.

[2007/HCI/I/7(modified)]

CJC · Discussion · Question 11

Answer

477/506477/506.

Scale the original sum and remove the first 21 terms

The infinite original sum is 66. The target numerator is three times the original, so the answer is 3[6−(6−5/22−2/23)]=3(5/22+2/23)=477/5063[6-(6-5/22-2/23)]=3(5/22+2/23)=477/506.

How did this part go?

Common mistakes

Apply the factor of three to both the infinite sum and the subtracted partial sum.

Read the hint