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What is the value of 1/3C2 + 1/4C2 + 1/5C2 + … + 1/100C2?

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What is the value of 1/3C2 + 1/4C2 + 1/5C2 + … + 1/100C2?

by Max@Math Revolution » Thu Mar 28, 2019 11:11 pm

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[GMAT math practice question]

What is the value of 1/3C2 + 1/4C2 + 1/5C2 + ... + 1/100C2?

A. 49/50
B. 51/50
C. 99/100
D. 101/100
E. 49/100
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by Max@Math Revolution » Sun Mar 31, 2019 5:28 pm
=>

nCr = { n(n-1)...(n-r+1) } / { 1*2*... *r }
So, nC2 = n(n-1)/2 and 1/nC2 = 2/{n(n-1)}.
Recall that 1/{k(k+1)} = 1/k - 1/(k+1).
So,
1/3C2 + 1/4C2 + 1/5C2 + ... + 1/100C2
= 2/(2*3) + 2/(3*4) + 2/(4*5) + ... + 2/(99*100)
= 2{ 1/(2*3) + 1/(3*4) + 1/(4*5) + ... + 1/(99*100) }
= 2{ ( 1/2 - 1/3 ) + ( 1/3 - 1/4 ) + ( 1/4 - 1/5 ) + ... + ( 1/99 - 1/100 ) }
= 2( 1/2 - 1/100 )
= 2 ( 50 / 100 - 1 / 100 )
= 2(49/100)
= 49/50

Therefore, the answer is A.
Answer: A
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