VJesus12 wrote: ↑

Fri May 14, 2021 2:05 am

For every integer \(k\) from \(1\) to \(10,\) inclusive the \(k \,th\) term of a certain sequence is given by \((-1)^{k+1}\cdot \dfrac1{2k}.\) If \(T\) is the sum of the first \(10\) terms in the sequence, then \(T\) is

A. Greater than \(2.\)

B. Between \(1\) and \(2.\)

C. Between \(\dfrac12\) and \(1.\)

D. Between \(\dfrac14\) and \(\dfrac12.\)

E. Less than \(\dfrac14.\)

Answer: D

Source: GMAT Prep

List some terms to see the pattern.

We get: T = 1/2 - 1/4 + 1/8 - 1/16 + . . .
Notice that we can rewrite this as T = (1/2 - 1/4) + (1/8 - 1/16) + . . .

When you start simplifying each part in brackets, you'll see a pattern emerge. We get...
T = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024

Now examine the last 4 terms: 1/16 + 1/64 + 1/256 + 1/1024
Notice that 1/64, 1/256, and 1/1024 are each less than 1/16
So, (1/16 + 1/64 + 1/256 + 1/1024) < (1/16 + 1/16 + 1/16 + 1/16)

Note: 1/16 + 1/16 + 1/16 + 1/16 = 1/4
So, we can conclude that 1/16 + 1/64 + 1/256 + 1/1024 = (a number less than 1/4)

Now start from the beginning: T = 1/4 + (1/16 + 1/64 + 1/256 + 1/1024)
= 1/4 + (a number less 1/4)
= A number less than 1/2
Of course, we can also see that T > 1/4
So, 1/4 < T < 1/2

Answer: D
Cheers,
Brent