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.