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18 Nov 2021, 05:45
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If \(a + \frac{a}{1 + 2} + \frac{a}{1 + 2 + 3} + ... + \frac{a}{1 + 2 + 3 + ... + 101} = 101\), then what is the value of \(a\) ?
A. \(50\)
B. \(51\)
C. \(52\)
D. \(55\)
E. \(100\)
A. \(50\)
B. \(51\)
C. \(52\)
D. \(55\)
E. \(100\)
18 Nov 2021, 05:45
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Official Solution:
If \(a + \frac{a}{1 + 2} + \frac{a}{1 + 2 + 3} + ... + \frac{a}{1 + 2 + 3 + ... + 101} = 101\), then what is the value of \(a\) ?
A. \(50\)
B. \(51\)
C. \(52\)
D. \(55\)
E. \(100\)
\(a + \frac{a}{1 + 2} + \frac{a}{1 + 2 + 3} + ... + \frac{a}{1 + 2 + 3 + ... + 101} = 101\);
Factor out \(a\): \(a(1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + ... + \frac{1}{1 + 2 + 3 + ... + 101}) = 101\);
In the denominators we have the sum of the first \(n\) positive integers, which equals to \(\frac{n(n+1)}{2}\).
Re-write the denominators according to the above: \(a(1 + \frac{1}{ \frac{2(2+1)}{2} } + \frac{1}{ \frac{3(3+1)}{2} } + ... + \frac{1}{ \frac{101(101+1)}{2} }) = 101\);
Simplify: \(a(1 + \frac{2}{2(2 +1 )} + \frac{2}{3(3+1)} + ... + \frac{2}{101(101+1)}) = 101\);
Factor out 2: \(2a(\frac{1}{2} + \frac{1}{2(2 +1 )} + \frac{1}{3(3+1)} + ... + \frac{1}{101(101+1)}) = 101\);
\(2a(\frac{1}{2} + \frac{1}{2*3} + \frac{1}{3*4} + ... + \frac{1}{101*102}) = 101\);
\(2a((1-\frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + ... + (\frac{1}{101} - \frac{1}{102})) = 101\);
Notice that everything in the brackets except the first and last number will cancel out and we'll get: \(2a(1 - \frac{1}{102}) = 101\);
\(2a(\frac{101}{102}) = 101\);
\(a=51\).
Answer: B
If \(a + \frac{a}{1 + 2} + \frac{a}{1 + 2 + 3} + ... + \frac{a}{1 + 2 + 3 + ... + 101} = 101\), then what is the value of \(a\) ?
A. \(50\)
B. \(51\)
C. \(52\)
D. \(55\)
E. \(100\)
\(a + \frac{a}{1 + 2} + \frac{a}{1 + 2 + 3} + ... + \frac{a}{1 + 2 + 3 + ... + 101} = 101\);
Factor out \(a\): \(a(1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + ... + \frac{1}{1 + 2 + 3 + ... + 101}) = 101\);
In the denominators we have the sum of the first \(n\) positive integers, which equals to \(\frac{n(n+1)}{2}\).
Re-write the denominators according to the above: \(a(1 + \frac{1}{ \frac{2(2+1)}{2} } + \frac{1}{ \frac{3(3+1)}{2} } + ... + \frac{1}{ \frac{101(101+1)}{2} }) = 101\);
Simplify: \(a(1 + \frac{2}{2(2 +1 )} + \frac{2}{3(3+1)} + ... + \frac{2}{101(101+1)}) = 101\);
Factor out 2: \(2a(\frac{1}{2} + \frac{1}{2(2 +1 )} + \frac{1}{3(3+1)} + ... + \frac{1}{101(101+1)}) = 101\);
\(2a(\frac{1}{2} + \frac{1}{2*3} + \frac{1}{3*4} + ... + \frac{1}{101*102}) = 101\);
\(2a((1-\frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + ... + (\frac{1}{101} - \frac{1}{102})) = 101\);
Notice that everything in the brackets except the first and last number will cancel out and we'll get: \(2a(1 - \frac{1}{102}) = 101\);
\(2a(\frac{101}{102}) = 101\);
\(a=51\).
Answer: B
22 Nov 2023, 00:19
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I got this in an adaptive test and tbh I had tried to simplify till the part where we can see it's the sum of a fractional series. But after that I just resorted to substituting 'a' values and guessing. The official solution described sounds ideal but I feel it is way too long to do during the exam, any other thumb rules or short cuts Bunuel?
