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Nice solution as usual, Frankenstein.Frankenstein wrote:Hi,
Tk = k(k+1)/2
From(1):
For Tn=6, n = 3 (odd)
For Tn=10, n = 4 (even)
Not sufficient
From(2):
T2n = (2n)(2n+1)/2 = n(2n+1).
2n+1 is always odd. So, for n(2n+1) to be even, n has to be even.
Sufficient
Hence, B
Hello Brent - i did not understand the solution - can you explain please how the formula becomes 1 + 2 + 3 + 4 + ...k = k(k+1)/2?Brent@GMATPrepNow wrote:Nice solution as usual, Frankenstein.Frankenstein wrote:Hi,
Tk = k(k+1)/2
From(1):
For Tn=6, n = 3 (odd)
For Tn=10, n = 4 (even)
Not sufficient
From(2):
T2n = (2n)(2n+1)/2 = n(2n+1).
2n+1 is always odd. So, for n(2n+1) to be even, n has to be even.
Sufficient
Hence, B
I thought I should point out to other viewers that Frankenstein is using a formula for finding the sum of all positive integers from 1 to k.
The formula goes like this: 1 + 2 + 3 + 4 + ...k = k(k+1)/2
Cheers,
Brent
The formula k(k+1)/2 is specifically for the sum of consecutive integers from 1 to k.subhakam wrote:Thank you Brent - this helps greatly.
Follow up question is - whether the below formula valid only for consecutive integers or for other evenly spaced sets as well where the difference between any 2 terms could be 5, 6 , -2 , 1/2 etc? What would be the formula for consecutive integers in decreasing order?
Lastly, the above does not hold valid (in my opinion) where some of the integers might be repeated in the set ? E.g. 1+2+3+3+3+4+4+5+5........+k ?
Many thanks once again!!
Subhakam
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