nguyendinhtuong wrote:
What is the value of \(A=\frac{1}{1 \times 2 \times 3}+\frac{1}{2 \times 3 \times 4}+...+\frac{1}{48 \times 49 \times 50}\)?
A. \(\frac{306}{1225}\)
B. \(\frac{637}{2550}\)
C. \(\frac{1175}{4704}\)
D. \(\frac{1199}{4800}\)
E. \(\frac{2497}{9996}\)
In these kind of questions our strategy will always be to break down each term into difference of two terms, such that alternate terms cancel out. In thsi case we can do it as follows
\(A=\frac{1}{2} * \frac{2}{1 \times 2 \times 3}+\frac{2}{2 \times 3 \times 4}+...+\frac{2}{48 \times 49 \times 50}\)
\(A=\frac{1}{2} * \frac{3-1}{1 \times 2 \times 3}+\frac{4-2}{2 \times 3 \times 4}+...+\frac{50-48}{48 \times 49 \times 50}\)
\(A=\frac{1}{2} * ( \frac{3}{1 \times 2 \times 3} - \frac{1}{1 \times 2 \times 3} + \frac{4}{2 \times 3 \times 4} - \frac{2}{2 \times 3 \times 4}+...+\frac{50}{48 \times 49 \times 50} - \frac{48}{48 \times 49 \times 50} )\)
\(A=\frac{1}{2} * ( \frac{1}{1 \times 2 } - \frac{1}{2 \times 3} + \frac{1}{2 \times 3 } - \frac{1}{3 \times 4}+...+\frac{1}{48 \times 49} - \frac{1}{49 \times 50} )\)
Now alternate terms will cancel out and we will be left with only two terms
\(A=\frac{1}{2} * ( \frac{1}{1 \times 2} - \frac{1}{49 \times 50} )\)
\(A=\frac{1}{2} * ( \frac{1225-1}{49 \times 50} )\)
\(A=\frac{1}{2} * ( \frac{1224}{49 \times 50} )\)
\(A=\frac{306}{1225}\)
So, answer will be A
Hope it helps!
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