Question:
The coefficient of $x$ in the expansion of $ (14+x)(1+2x)(1+3x)....(1+100x) $ is
The coefficient of $x$ in the expansion of $ (14+x)(1+2x)(1+3x)....(1+100x) $ is
Updated On: May 22, 2024
- 5050
- 10100
- 5151
- 4950
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The Correct Option is A
Solution and Explanation
The coefficient of $ x $ in the expansion of
$ (1+x)(1+2x)(1+3x)....(1+100x) $
$=1+2+3+....+100 $
$=\frac{100(100+1)}{2}=50\times 101=5050 $
$ (1+x)(1+2x)(1+3x)....(1+100x) $
$=1+2+3+....+100 $
$=\frac{100(100+1)}{2}=50\times 101=5050 $
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Concepts Used:
Binomial Theorem
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
Properties of Binomial Theorem
- The number of coefficients in the binomial expansion of (x + y)n is equal to (n + 1).
- There are (n+1) terms in the expansion of (x+y)n.
- The first and the last terms are xn and yn respectively.
- From the beginning of the expansion, the powers of x, decrease from n up to 0, and the powers of a, increase from 0 up to n.
- The binomial coefficients in the expansion are arranged in an array, which is called Pascal's triangle. This pattern developed is summed up by the binomial theorem formula.