Respuesta :
Answer with explanation:
The expansion of
[tex](1+x)^n=1 + nx +\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+......[/tex]
where,n is a positive or negative , rational number.
Where, -1< x < 1
Expansion of
[tex](1-x^2)^{-5}=1-5 x^2+\frac{(5)\times (6)}{2!}x^4-\frac{5\times 6\times 7}{3!}x^6+\frac{5\times 6\times 7\times 8}{4!}x^8-\frac{5\times 6\times 7\times 8\times 9}{5!}x^{10}+\frac{5\times 6\times 7\times 8\times 9\times 10}{6!}x^{12}+....[/tex]
Coefficient of [tex]x^{12}[/tex] in the expansion of [tex](1-x^2)^{-5}[/tex] is
[tex]=\frac{5\times 6\times 7\times 8\times 9\times 10}{6!}\\\\=\frac{15120}{6\times 5 \times 4\times 3 \times 2 \times 1}\\\\=\frac{151200}{720}\\\\=210[/tex]
⇒As the expansion [tex](1-x^2)^{-5}[/tex] contains even power of x , so there will be no term containing [tex]x^{17}[/tex].
