Question

Expand x^n to the 4th term using binomial expansion

Answer 1

To expand \( (x + y)^n \) using the binomial expansion, we can use the Binomial Theorem, which states:

\[
(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k
\]

where \( \binom{n}{k} \) is the binomial coefficient calculated as \( \frac{n!}{k!(n-k)!} \).

To specifically find the first four terms of the expansion, we can set \( y = 1 \). Therefore, the expansion becomes:

\[
(x + 1)^n = \sum_{k=0}^{3} \binom{n}{k} x^{n-k} \cdot 1^k
\]

Calculating the first four terms for \( k = 0, 1, 2, \) and \( 3 \):

1. For \( k = 0 \):

\[
\binom{n}{0} x^{n-0} = 1 \cdot x^n = x^n
\]

2. For \( k = 1 \):

\[
\binom{n}{1} x^{n-1} = n \cdot x^{n-1}
\]

3. For \( k = 2 \):

\[
\binom{n}{2} x^{n-2} = \frac{n(n-1)}{2} x^{n-2}
\]

4. For \( k = 3 \):

\[
\binom{n}{3} x^{n-3} = \frac{n(n-1)(n-2)}{6} x^{n-3}
\]

Thus, the first four terms of the binomial expansion of \( (x + 1)^n \) are:

\[
x^n + n x^{n-1} + \frac{n(n-1)}{2} x^{n-2} + \frac{n(n-1)(n-2)}{6} x^{n-3}
\]

This provides the fourth term in the expansion as:

\[
\frac{n(n-1)(n-2)}{6} x^{n-3}
\]

So, the full expansion up to the fourth term is:

\[
(x + 1)^n = x^n + n x^{n-1} + \frac{n(n-1)}{2} x^{n-2} + \frac{n(n-1)(n-2)}{6} x^{n-3} + \ldots
\]