Question

Binomial expansion...help me pls​

Answer 1

By the binomial theorem,

`\displaystyle \left(1+\frac xk\right)^n = \sum_(i=0)^n \binom ni \left(\frac xk\right)^n`

The
`x^2` term occurs for i = 2, and the
`x^3` term for i = 3. If the coefficient of the squared term is 3 times larger than the coefficient of the cubed term, then

`\dbinom n2 \frac1{k^2} = 3\dbinom n3\frac1{k^3}`

Solve for k :

`\dbinom n2 \frac1{k^2} = 3 \dbinom n3 \frac1{k^3} \\\\ (n!)/(2!(n-2)!) \frac1{k^2} = (3n!)/(3!(n-3)!) \frac1{k^3} \\\\ (n!)/(2(n-2)(n-3)!) \frac1{k^2} = (3n!)/(6(n-3)!) \frac1{k^3} \\\\ (1)/((n-2)k^2) = (1)/(k^3)\\\\ k^3 = (n-2)k^2 \\\\ k^3 - (n-2)k^2 = 0 \\\\ k^2 (k - (n - 2)) = 0`

Since k ≠ 0, it follows that

k - (n - 2) = 0

or

k = n - 2

as required.