Question

The sum of the series $^4{C_0}{ + ^5}{C_1}x{ + ^6}{C_2}{x^2}{ + ^7}{C_3}{x^3} + \cdots \,{\text{to}}\,\infty$ is
A. ${\left( {1 - x} \right)^{ - 4}}$
B. $\dfrac{1}{{{{\left( {1 - x} \right)}^5}}}$
C. ${\left( {1 + x} \right)^{ - 5}}$
D.None of these

Answer 1

Hint : To solve this question first we should know that ${\left( {1 - x} \right)^{ - n}}$ expansion is $^{n - 1}{C_0}{ + ^n}{C_1}x{ + ^{n + 1}}{C_2}{x^2}{ + ^{n + 2}}{C_3}{x^3} + \cdots$
comparing the coefficients of the terms of the standard series with that given in question we solve the given problem.

Complete step-by-step answer :
Given, series is $^4{C_0}{ + ^5}{C_1}x{ + ^6}{C_2}{x^2}{ + ^7}{C_3}{x^3} + \cdots \,{\text{to}}\,\infty$ .
As, we know the expansion of the ${\left( {1 - x} \right)^{ - n}}$ is given by $^{n - 1}{C_0}{ + ^n}{C_1}x{ + ^{n + 1}}{C_2}{x^2}{ + ^{n + 2}}{C_3}{x^3} + \cdots$ .
If we compare equation $^4{C_0}{ + ^5}{C_1}x{ + ^6}{C_2}{x^2}{ + ^7}{C_3}{x^3} + \cdots \,{\text{to}}\,\infty$ and $^{n - 1}{C_0}{ + ^n}{C_1}x{ + ^{n + 1}}{C_2}{x^2}{ + ^{n + 2}}{C_3}{x^3} + \cdots$ .
Then, $n - 1 = 4 \Rightarrow n = 4 + 1 = 5$
Therefore, n will be equal to 5.
So, by this we can conclude that $^4{C_0}{ + ^5}{C_1}x{ + ^6}{C_2}{x^2}{ + ^7}{C_3}{x^3} + \cdots \,{\text{to}}\,\infty$ sums up as ${\left( {1 - x} \right)^{ - 4}}$ .
So, the correct answer is “${\left( {1 - x} \right)^{ - 4}}$ ”.

Note : The Binomial Theorem is the process of extending an expression to some finite power that has been elevated. A binomial theorem is a strong expansion instrument that has Algebra application, likelihood, etc.