Source: GMAT Prep
For every integer \(k\) from 1 to 10, inclusive the "\(k\)"th term of a certain sequence is given by \((-1)^{(k+1)}\cdot \left(\frac{1}{2^k}\right)\) 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
The OA is D
For every integer \(k\) from 1 to 10, inclusive the "k&
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T = 1/2 - 1/4 + 1/8 - 1/16 + . . .BTGmoderatorLU wrote:Source: GMAT Prep
For every integer \(k\) from 1 to 10, inclusive the "\(k\)"th term of a certain sequence is given by \((-1)^{(k+1)}\cdot \left(\frac{1}{2^k}\right)\) 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
The OA is D
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
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$$\left(-1\right)^{k+1}=\left(\frac{1}{2^k}\right)$$
Let k = 1
$$\left(-1\right)^2\cdot\frac{1}{2^1}=1\cdot\frac{1}{2}=\frac{1}{2}=T_1$$
Let k = 2
$$\left(-1\right)^3\cdot\frac{1}{2^2}=-1\cdot\frac{1}{4}=-\frac{1}{4}=T_2$$
$$T=T_1+T_2+--------+T_n$$
$$T=\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{8}-\frac{1}{16}\right)+\left(\frac{1}{32}-\frac{1}{64}\right)+\left(\frac{1}{128}-\frac{1}{256}\right)-\left(\frac{1}{512}-\frac{1}{1024}\right)$$
$$First\ term\ \left(a\right)=\frac{1}{2},\ ratio\ =\ -\frac{1}{2}$$
$$Sum\ of\ G.p\ where\ r<1=\frac{a\left(1-r^n\right)}{1-r}=\frac{a}{\ 1-r}$$
$$\frac{\frac{1}{2}}{1-\left(-\frac{1}{2}\right)}=\frac{1}{3}$$
Therefore,
$$sum\ of\ sequence\ is\ close\ to\ \frac{1}{3}and\ cannot\ be\ greater\ than\ \frac{1}{3\ }$$
$$Answer\ is\ Option\ D$$
Let k = 1
$$\left(-1\right)^2\cdot\frac{1}{2^1}=1\cdot\frac{1}{2}=\frac{1}{2}=T_1$$
Let k = 2
$$\left(-1\right)^3\cdot\frac{1}{2^2}=-1\cdot\frac{1}{4}=-\frac{1}{4}=T_2$$
$$T=T_1+T_2+--------+T_n$$
$$T=\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{8}-\frac{1}{16}\right)+\left(\frac{1}{32}-\frac{1}{64}\right)+\left(\frac{1}{128}-\frac{1}{256}\right)-\left(\frac{1}{512}-\frac{1}{1024}\right)$$
$$First\ term\ \left(a\right)=\frac{1}{2},\ ratio\ =\ -\frac{1}{2}$$
$$Sum\ of\ G.p\ where\ r<1=\frac{a\left(1-r^n\right)}{1-r}=\frac{a}{\ 1-r}$$
$$\frac{\frac{1}{2}}{1-\left(-\frac{1}{2}\right)}=\frac{1}{3}$$
Therefore,
$$sum\ of\ sequence\ is\ close\ to\ \frac{1}{3}and\ cannot\ be\ greater\ than\ \frac{1}{3\ }$$
$$Answer\ is\ Option\ D$$
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We are given that for every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) x (1/2^k). We must determine the sum of the first 10 terms in the sequence. Before calculating the sum, we should recognize that the answer choices are provided as ranges of values, rather than as exact values. Thus, we might not need to calculate the total of the 10 terms to determine an answer. Perhaps we can uncover a pattern to help us find the answer. Let's start by listing the first four terms.BTGmoderatorLU wrote:Source: GMAT Prep
For every integer \(k\) from 1 to 10, inclusive the "\(k\)"th term of a certain sequence is given by \((-1)^{(k+1)}\cdot \left(\frac{1}{2^k}\right)\) 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
The OA is D
k = 1:
(-1)^(1+1) x (1/2^1)
(-1)^2 x 1/2
1 x 1/2 = 1/2
k = 2:
(-1)^(2+1) x (1/2^2)
(-1)^3 x 1/4
-1 x 1/4 = -1/4
k = 3:
(-1)^(3+1) x (1/2^3)
(-1)^4 x 1/8
1 x 1/8 = 1/8
k = 4:
(-1)^(4+1) x (1/2^4)
(-1)^5 x 1/16
-1 x 1/16 = -1/16
Recall that we are trying to estimate the value of T = 1/2 + (-1/4) + 1/8 + (-1/16) + ... until we have 10 terms. In other words, T = 1/2 - 1/4 + 1/8 - 1/16 + ... until there are 10 terms.
We should notice that the absolute values of the terms are getting smaller:
|1/2|>|-1/4|>|1/8|>|-1/16|.
Notice that starting from the first term of 1/2, we are subtracting something less than 1/2 (notice that 1/4 < 1/2) but then adding back something even less (notice 1/8 < 1/4), and the process continues. Thus, because ½ and -1/4 are the largest term and the smallest term, respectively, in our set, the sum will never fall below ¼ or exceed ½.
Thus, we conclude that T is greater than 1/4 but less than 1/2.
Answer: D
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