The coeficent of x²⁰ in the expansion of (1+x²)⁴⁰(x²+2+1/x²)^-5 is???

 

pls also explain ur answer

This can be done by expanding (x^2+2+1/x^2)^5
you will get the cofficient of some terms and just find the corresponding coefficients from the second
term ie (1+x^2)^40
you will get the corresponding terms for each term you get from expanding (x^2+2+1/x^2)^5
for eg : corresponding term for 1/x^10 will be x^30 and for x^10 will be x^10
except of the term containing 1/x^5
just multiply the corresponding coefficients and evaluate the result by adding them all
the result would be :

⁴⁰c₁₄ + ⁴⁰c₁₃ + 15⁴⁰c₁₂/2 + 5⁴⁰c₁₁/4 + 165⁴⁰c₁₀/16
+ 481⁴⁰c₉/32 + 285⁴⁰c₈/16 + 45⁴⁰c₇/4 + 5⁴⁰c₆/2 + 5⁴⁰c₅/2 + ⁴⁰c₄

 

hii

 

let me try to solve this ..

 

 (1+x²)⁴⁰(x²+2+1/x²)^-5  = ( x +  1/x )^-2.5.(1+x²)⁴⁰

 

                                             = [(1+x²)⁴⁰(1+x²)^-10].x¹⁰

 

                                             = [(1+x²)³⁰].x¹⁰

 

So, now we need just the coeff of x¹⁰ in [(1+x²)³⁰]

 

which is ³⁰c_{5 .}

 

cheers