Question:
$ \frac{^{8}{{C}_{0}}}{6}{{-}^{8}}{{C}_{1}}{{+}^{8}}{{C}_{2}}\cdot 6{{-}^{8}}{{C}_{3}}{{.6}^{2}}+....{{+}^{8}}{{C}_{8}}{{\cdot 6}^{7}} $ is equal to
$ \frac{^{8}{{C}_{0}}}{6}{{-}^{8}}{{C}_{1}}{{+}^{8}}{{C}_{2}}\cdot 6{{-}^{8}}{{C}_{3}}{{.6}^{2}}+....{{+}^{8}}{{C}_{8}}{{\cdot 6}^{7}} $ is equal to
Updated On: Jun 23, 2024
- $ 0 $
- $ {{6}^{7}} $
- $ {{6}^{8}} $
- $ \frac{{{5}^{8}}}{6} $
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The Correct Option is D
Solution and Explanation
$ \frac{^{8}{{C}_{0}}}{6}{{-}^{8}}{{C}_{1}}{{+}^{8}}{{C}_{2}}.6{{-}^{8}}{{C}_{3}}{{6}^{2}}+....{{+}^{8}}{{C}_{8}}{{6}^{7}} $
$ =\frac{1}{6}{{[}^{3}}{{C}_{0}}-{{6}^{8}}{{C}_{1}}+{{6}^{2}}{{\,}^{8}}{{C}_{2}}-{{6}^{3}}{{\,}^{8}}{{C}_{3}}+....+{{6}^{8}}{{\,}^{6}}{{C}_{8}}] $
$ =\frac{1}{6}[{{(1-6)}^{8}}]=\frac{{{5}^{8}}}{6} $
$ =\frac{1}{6}{{[}^{3}}{{C}_{0}}-{{6}^{8}}{{C}_{1}}+{{6}^{2}}{{\,}^{8}}{{C}_{2}}-{{6}^{3}}{{\,}^{8}}{{C}_{3}}+....+{{6}^{8}}{{\,}^{6}}{{C}_{8}}] $
$ =\frac{1}{6}[{{(1-6)}^{8}}]=\frac{{{5}^{8}}}{6} $
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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.