Question

lf $y=(x^2-1{)}^n$, then the value of $(x^2-1)y_{n+2}+2x{y}_{n+1}$ is

• A. $(x^2+1)y_n$
• B. $(x^2-1)y_n$
• C. $n(n+1)y_n$
• D. $n(n^2+1)y_n$

Answer 1

The correct option is C $n(n+1)y_n$

Given that:

$y=(x^2-1{)}^n$

To Find:

$(x^2-1)y_{n+2}+2x{y}_{n+1}=?$

Solution:

$y=(x^2-1{)}^n$

On diffrentiation:

or, $y_1=2nx(x^2-1{)}^{n-1}$

or, $(x^2-1)y_1=2nxy$

or, $(x^2-1)y_2+2x{y}_1=2nx{y}_1+2ny$

or, $(x^2-1)y_2+2x(1-n)y_1=2ny$

Differentiating $n$ times following Leibnitz's theorem,

$(x^2-1)y_{n+2}+2nx{y}_{n+1}+n(n-1)y_n+2x(1-n)y_{n+1}+2n(1-n)y_n=2n{y}_n$

or, $(x^2-1)y_{n+2}+(2nx+2x-2nx)y_{n+1}=(2n-2n+2n^2-n^2+n)y_n$

or, $(x^2-1)y_{n+2}+2x{y}_{n+1}=n(n+1)y_n$

Hence, C is the correct option.