Fun with series 2

Algebra Level 4

The value of

2 1 × 2 × 3 − 2 2 × 3 × 4 − ⋯ − 2 19 × 20 × 21 − 2 20 × 21 × 22 \dfrac{2}{1\times2\times3}-\dfrac{2}{2\times3\times4}-\cdots-\dfrac{2}{19\times20\times21}-\dfrac{2}{20\times21\times22}

is m n \dfrac{m}{n} where m m and n n are co-prime. Find m + n m+n .


The answer is 90.

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Tom Zhou by Tom Zhou , 21, Canada

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2 solutions

2/(n-1)n(n+1) = - 1/(n-1) + 2/n - 1/(n+1)
-2/(n-1)n(n+1) = 1/(n-1) - 2/n + 1/(n+1)
- 2/2x3x4 - 2/3x4x5 - ............ - 2/20x21x22 = 1/2 - 2/3 + 1/4 + 1/3 - 2/4 + 1/5 + ......... + 1/20 - 2/21 + 1/22 =
1/2 - 1/3 - 1/21 + 1/22 + 2/1x2x3 = 1/2 - 1/3 - 1/21 + 1/22 - 1/1 + 2/2 - 1/3 =
1/2 + 1/22 - 2/3 - 1/21 = 6/11 - 5/7 = - 13/77
13+77 = 90

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Did it a little differently (and more hard work).

I rewrote the series as 2/3 - (2/[n(n+1)(n+2)]).

Yours is quite clever

Ceesay Muhammed - 6 years, 6 months ago

done in exactly that way

Priyesh Pandey - 6 years, 8 months ago

i wrote 2 as 3-1,4-2,5-3 ...... and then i got a telescopic series and i was left with 4 terms 1/1x2 -1/2x3 -1/2x3 -1/21x22

Shekhar Patole - 6 years, 8 months ago

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Brilliant!

Edwin Hughes - 6 years, 7 months ago
William Chau
Jan 26, 2015

Since

2/(1 * 2 * 3)-2/(2 * 3 * 4)-...-2/(19 * 20 * 21)-2/(20 * 21 * 22)

= 2 * 2/(1 * 2 * 3)-[2/(1 * 2 * 3)+2/(2 * 3 * 4)+...+2/(19 * 20 * 21)+2/(20 * 21 * 22)]

= 2/3-[1/(1 * 2)-1/(2 * 3)+1/(2 * 3)-1/(3 * 4)+...+1/(19 * 20)-1/(20 * 21)+1/(20 * 21)-1/(21 * 22)]

= 2/3-[1/2-1/(21 * 22)]

= 13/77,

so m = 13, n = 77, and m+n = 90.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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