Question

Find the coefficient of x^2 in the expansion of (1-x-x^2)*(x+1/x)^10

Answer 1

Answer:

-42

I hope this helps.

Step-by-step explanation:

Let f(x) = (1-x-x²) ( x + 1/x )¹⁰.

Multiplying by x¹⁰, the coefficient of x² in f(x) is just

coeff of x¹² in x¹⁰ f(x)

= coeff of x¹² in ( 1 - x - x² ) ( x² + 1 )¹⁰

= ( coeff of x¹² in ( x² + 1 )¹⁰ )  -  ( coeff of x¹² in x ( x² + 1 )¹⁰ )

   -  ( coeff of x¹² in x² ( x² + 1 )¹⁰ )

= ( coeff of x¹² in ( x² + 1 )¹⁰ )  -  ( coeff of x¹¹ in ( x² + 1 )¹⁰ )

   -  ( coeff of x¹⁰ in ( x² + 1 )¹⁰ )

Now since ( x² + 1 )¹⁰ only involves powers of x², the coefficient of x¹¹ is 0.

We are then left with:

coeff of x² in f(x)

=  ( coeff of x¹² in ( x² + 1 )¹⁰ )  -  ( coeff of x¹⁰ in ( x² + 1 )¹⁰ )

=  ( coeff of x⁶ in ( x + 1 )¹⁰ )  -  ( coeff of x⁵ in ( x + 1 )¹⁰ )

= $\displaystyle\binom{10}6 - \binom{10}5$

= $\displaystyle\binom{10}{4}\left(1-\tfrac65\right)$

= $\displaystyle-\frac{1}{5}\times\frac{10\times9\times8\times7}{4\times3\times2}$

= -42