If N is a positive integer and TN is the
sum of all the positive integers from 1 to
N, inclusive, is N even?
(1) TN is even.
(2) T2N is even.
Note : the N & 2N are subscript.
even odd
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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
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
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
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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
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
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
Thank you very much!
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1 + 2 + 3 + 4 + ...+k = k(k+1)/2
So, for example, the sum 1+2+3+4+.....+39+40 = (40)(41)/2 = 820
Similarly, 1+2+3+4+.....+58+59 = (59)(60)/2 = 1770
Does that answer your question?
Cheers,
Brent
So, for example, the sum 1+2+3+4+.....+39+40 = (40)(41)/2 = 820
Similarly, 1+2+3+4+.....+58+59 = (59)(60)/2 = 1770
Does that answer your question?
Cheers,
Brent
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
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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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
Having said that, we can often make some adjustments to our sum in order to apply the formula.
For example:
7+14+21+28+....+203+210 = 7(1+2+3+4+....29+30)
= 7[(30)(31)/2]
= 7[465]
= 3255
Cheers,
Brent