Question

Coefficient of x^2 y^20 in expansion of (x-2y^4)^7

Answer 1

The coefficient of
`\(x^2y^(20)\)` in the expansion of
`\((x-2y^4)^7\)` is -672, obtained by applying the binomial theorem with k = 5 and evaluating the expression step by step.

Let's break down the calculation step by step.

The general term in the expansion of
`\((x-2y^4)^7\)` is given by the binomial theorem:

`\[T_k = \binom{7}{k} x^(7-k) (-2y^4)^k\]`

Now, for the term corresponding to
`\(x^2y^(20)\)`, we need to find the value of \(k\) that satisfies two conditions:

1. The power of x: 7 - k = 2 (because we want
`\(x^2\))`

2. The power of y: 4k = 20 (because we want
`\(y^(20)\))`

Solving 7 - k = 2 gives k = 5. Substituting this into 4k = 20 confirms that it's satisfied.

Now, plug k = 5 into the general term formula:

`\[T_5 = \binom{7}{5} x^(7-5) (-2y^4)^5\]`

Simplify this expression:

`\[T_5 = \binom{7}{5} x^2 (-32y^(20))\]`

Now, calculate the binomial coefficient:

`\[\binom{7}{5} = (7!)/(5!(7-5)!) = (7 * 6)/(2) = 21\]`

Substitute this back into the expression:

`\[T_5 = 21 * x^2 * (-32y^(20))\]`

Finally, calculate the coefficient:

`\[21 * (-32) = -672\]`

So, the coefficient of
`\(x^2y^(20)\)` in the expansion of
`\((x-2y^4)^7\) is \(-672\).`