GeminiHeat wrote:
For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by \((-1)^{(k+1)}*(\frac{1}{2^k})\). 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 1/2 and 1
D. Between 1/4 and 1/2
E. Less than 1/4
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 + 1/1024Notice 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 = 1/4So, 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.