Question: If the number of terms in $\left( x + 1 + \frac{1}{x} \right)^{n}$ (n Î I<sup>+</sup>) is 401 then...

Question

If the number of terms in $\left( x + 1 + \frac{1}{x} \right)^{n}$ (n Î I⁺) is 401 then n is greater than-

Answer 1

$\left( x + 1 + \frac{1}{x} \right)^{n}$ = $\frac{(1 + x + x^{2})^{n}}{x^{n}}$

(1 + x + x²)ⁿ is the form a₀ + a₁x + a₂x² +.....+ a_2nx²ⁿ

which contains 2n + 1 terms

2n + 1 = 401 ̃ n = 200

Question: If the number of terms in \(\left( x + 1 + \frac{1}{x} \right)^{n}\) (n Î I<sup>+</sup>) is 401 then...