Question

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

Answer 1

The Binomial Theorem states that for any integer \( n \):

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

To expand \( (x + 1)^n \) to the 4th term, we want to identify the first four terms of the expansion. The binomial coefficients are defined as:

\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\]

We will consider \( y = 1 \):

1. The first term (when \( k = 0 \)):
\[
\binom{n}{0} x^n \cdot 1^0 = 1 \cdot x^n = x^n
\]

2. The second term (when \( k = 1 \)):
\[
\binom{n}{1} x^{n-1} \cdot 1^1 = n \cdot x^{n-1}
\]

3. The third term (when \( k = 2 \)):
\[
\binom{n}{2} x^{n-2} \cdot 1^2 = \frac{n(n-1)}{2} x^{n-2}
\]

4. The fourth term (when \( k = 3 \)):
\[
\binom{n}{3} x^{n-3} \cdot 1^3 = \frac{n(n-1)(n-2)}{6} x^{n-3}
\]

Combining all of these, the expansion of \( (x + 1)^n \) up to the 4th term is:

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

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

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