Question

The sum of the coefficients in the expansion of ${\left(5x-4y\right)}^n$, where $n$ is a positive integer is

• A. $0$
• B. $n$
• C. $1$
• D. $-1$

Answer 1

The correct option is C

$1$

Explanation for the correct answer:

The given expression ${\left(5x-4y\right)}^n$ can be expanded by using the Binomial theorem as

${\left(5x-4y\right)}^n={}^n{C}_0{\left(5x\right)}^n+{}^n{C}_1{\left(5x\right)}^4\left(-4y\right)+{}^n{C}_2{\left(5x\right)}^3{\left(-4y\right)}^2+.....+{}^n{C}_n{\left(-4y\right)}^n$

In the above expression if we substitute $x=y=1$, we get we get the sum of their coefficients. Thus,

${\left(5-4\right)}^n={}^n{C}_0{\left(5\right)}^n+{}^n{C}_1{\left(5\right)}^4\left(-4\right)+{}^n{C}_2{\left(5\right)}^3{\left(-4\right)}^2+....+{}^n{C}_n{\left(-4\right)}^n$

${}^n{C}_0{\left(5\right)}^n+{}^n{C}_1{\left(5\right)}^4\left(-4\right)+{}^n{C}_2{\left(5\right)}^3{\left(-4\right)}^2+....+{}^n{C}_n{\left(-4\right)}^n$ is the sum of all coefficients of the terms in the expansion.

$\therefore$Sum of all coefficients $={\left(5-4\right)}^n$

$=1^n$ ($\because$ any power of $1$ is $1$)

$=1$

Hence, the sum of coefficients of the terms in expansion of ${\left(5x-4y\right)}^n$ is $1$.

Hence, option C is the correct answer.