mmcooley33 wrote:
3/4, 5/36, 7/144
The numbers above form a sequence,t1,t2, and t3 , which is defined by tm = (1 / m^2) - (1/m+1^2)for all positive integers m. Is the sum of the first J terms of the sequence greater than 63/64?
1. j>8
2. j<16
In such kind of questions there is always a pattern in terms or/and in the sum of the terms.
Given: \(t_m=\frac{1}{m^2}-\frac{1}{(m+1)^2}\). So:
\(t_1=\frac{1}{1^2}-\frac{1}{(1+1)^2}=1-\frac{1}{2^2}\);
\(t_2=\frac{1}{2^2}-\frac{1}{(2+1)^2}=\frac{1}{2^2}-\frac{1}{3^2}\);
\(t_3=\frac{1}{3^2}-\frac{1}{(3+1)^2}=\frac{1}{3^2}-\frac{1}{4^2}\);
...
You should notice that if we have as sum of first 2 terms then every thing but the 1 from \(t_1\) and the last part from \(t_2\) (1/3^2=1/(2+1)^2) will cancel out, so \(sum_2=1-\frac{1}{(2+1)^2}\). The same if we sum first 3 terms: only 1 minus the last part of \(t_3\) (1/4^2=1/(3+1)^2) will remain, \(sum_3=1-\frac{1}{(3+1)^2}\). So if we sum first \(j\) terms the the sum will equal to \(1-\frac{1}{(j+1)^2}\).
Question: is \(Sum_j=1-\frac{1}{(j+1)^2}>\frac{63}{64}\) --> is \((j+1)^2>64\)--> is \(j>7\)?
(1) j>8. Sufficient.
(2) j<16. Not sufficient.
Answer: A.