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

## For every integer \(k\) from \(1\) to \(10,\) inclusive the \(k \,th\) term of a certain sequence is given by

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List some terms to see the pattern.VJesus12 wrote: ↑Fri May 14, 2021 2:05 amFor 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

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