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07 Jun 2021, 08:31
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B
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The sequence \(s_1\), \(s_2\), \(s_3\), ..., \(s_n\), ... is such that \(s_n=2^{n-1}\) for all integers \(n\geq{1}\). What is the sum of the first 10 terms of the sequence ?
A. 2^9 - 1
B. 2^10 - 1
C. 2^13 - 1
D. 2^14 - 1
E. 2^16 - 1
M37-57
A. 2^9 - 1
B. 2^10 - 1
C. 2^13 - 1
D. 2^14 - 1
E. 2^16 - 1
M37-57
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29 Nov 2021, 04:17
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Official Solution:
The sequence \(s_1\), \(s_2\), \(s_3\), ..., \(s_n\), ... is such that \(s_n=2^{n-1}\) for all integers \(n\geq{1}\). What is the sum of the first 10 terms of the sequence ?
A. \(2^9 - 1\)
B. \(2^{10} - 1\)
C. \(2^{13} - 1\)
D. \(2^{14} - 1\)
E. \(2^{16} - 1\)
\(n_{th}\) term of the sequence is given by\(s_n=2^{n-1}\), so: \(s_1=2^{1-1}=2^0\), \(s_2=2^{2-1}=2^1\), \(s_3=2^{3-1}=2^2\), ..., \(s_{10}=2^{10-1}=2^9\).
We should find the value of \(2^0 + 2^1 + 2^2 + ... + 2^9\)
\(2^0+2^1=3=2^2-1\) (the sum of first and the second terms);
\((2^2-1)+2^2=2*2^2-1=2^3-1\) (the sum of the previous 2 terms and the third term);
\((2^3-1)+2^3=2*2^3-1=2^4-1\) (the sum of the previous 3 terms and the fourth term);
...
The same will continue and finally we'll get \((2^9-1)+2^9=2*2^9-1=2^{10}-1\), the sum of the previous 9 terms and the 10th term.
OR: we can identify that the sequence is a geometric progression.
Sum of the terms of geometric progression is given by: \(Sum=\frac{a*(r^{n}-1)}{r-1}\), where \(a\) is the first term, \(n\) is the number of terms and \(r\) is a common ratio \(>1\). So:
\(2^0 + 2^1 + 2^2 + ... + 2^9=\frac{2^0*(2^{10}-1)}{2-1}=2^{10}-1\).
Answer: B
The sequence \(s_1\), \(s_2\), \(s_3\), ..., \(s_n\), ... is such that \(s_n=2^{n-1}\) for all integers \(n\geq{1}\). What is the sum of the first 10 terms of the sequence ?
A. \(2^9 - 1\)
B. \(2^{10} - 1\)
C. \(2^{13} - 1\)
D. \(2^{14} - 1\)
E. \(2^{16} - 1\)
\(n_{th}\) term of the sequence is given by\(s_n=2^{n-1}\), so: \(s_1=2^{1-1}=2^0\), \(s_2=2^{2-1}=2^1\), \(s_3=2^{3-1}=2^2\), ..., \(s_{10}=2^{10-1}=2^9\).
We should find the value of \(2^0 + 2^1 + 2^2 + ... + 2^9\)
\(2^0+2^1=3=2^2-1\) (the sum of first and the second terms);
\((2^2-1)+2^2=2*2^2-1=2^3-1\) (the sum of the previous 2 terms and the third term);
\((2^3-1)+2^3=2*2^3-1=2^4-1\) (the sum of the previous 3 terms and the fourth term);
...
The same will continue and finally we'll get \((2^9-1)+2^9=2*2^9-1=2^{10}-1\), the sum of the previous 9 terms and the 10th term.
OR: we can identify that the sequence is a geometric progression.
Sum of the terms of geometric progression is given by: \(Sum=\frac{a*(r^{n}-1)}{r-1}\), where \(a\) is the first term, \(n\) is the number of terms and \(r\) is a common ratio \(>1\). So:
\(2^0 + 2^1 + 2^2 + ... + 2^9=\frac{2^0*(2^{10}-1)}{2-1}=2^{10}-1\).
Answer: B
General Discussion
07 Jun 2021, 12:04
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Bunuel
S1 = 1
S2 = 2
S3 = 4
S4 = 8
If you notice all these terms are in GP.
So, \(\frac{a . r^n - 1}{r - 1}\) = \(2^{10} - 1\) -(B), IMO!
