Bunuel wrote:
SOLUTION
The sequence s1, s2, s3, ..., sn, ... is such that \(S_n=\frac{1}{n} - \frac{1}{n+1}\) for all integers \(n\geq{1}\). If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?
Given: \(s_n=\frac{1}{n}-\frac{1}{n+1}\) for \(n\geq{1}\). So:
\(s_1=1-\frac{1}{2}\);
\(s_2=\frac{1}{2}-\frac{1}{3}\);
\(s_3=\frac{1}{3}-\frac{1}{4}\);
...
If you sum the above 3 terms you'll get: \(s_1+s_2+s_3=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})=1-\frac{1}{4}\) (everything but the first and the last numbers will cancel out). So the sum of first \(k\) terms is fgiven by the formula \(sum_k=1-\frac{1}{k+1}\).
Question: is \(sum_k=1-\frac{1}{k+1}>\frac{9}{10}\)? --> is \(\frac{k}{k+1}>\frac{9}{10}\)? --> is \(k>9\)?
(1) k > 10. Sufficient.
(2) k < 19. Not sufficient.
Answer: A.
Hi
Bunuel I understood your point but I approached the problem in the following way :
Point 1: I saw the word “Sequence” — that’s why I have written the formula of Sum of n integers as :
Sum of first n integers in s sequence =
n/2( 1st term + Last term )
We have to prove that : Sum of 1st K terms in the sequence is > 9/10 i.e greater than 0.9
1) as per the first statement , K>10
So I took k’s value as 11 to check whether the sun is greater than 0.9 or not ,
Sum (11 integers ) = 11/2{ 1/2 + [(1/11)-1/12] }
Which gives something around 5.6
So , it’s not true
Again I took k =100
Sum (100 integers ) =100/2{1/2 + [1/100-1/101]}
The answer is something around 55
Now it’s true
That’s why A is not sufficient
Plz guide me regarding this approach for which I was pretty confident initially
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