Question

The term independent of x in the binomial expansion of \((x+\frac{2}{x^3})^{20}\) is

• A. \(\begin{pmatrix} 20 \\ 5 \end{pmatrix}2^{15}\)
• B. \(\begin{pmatrix} 20 \\ 15 \end{pmatrix}2^{10}\)
• C. \(\begin{pmatrix} 20 \\ 10 \end{pmatrix}2^{5}\)
• D. \(\begin{pmatrix} 20 \\ 10 \end{pmatrix}2^{10}\)
• E. \(\begin{pmatrix} 20 \\ 5 \end{pmatrix}2^{5}\)

Answer 1

Answer: (E)

We are asked to find the term independent of \( x \) in the binomial expansion of \( \left(x + \frac{2}{x^3}\right)^{20} \).

The binomial expansion of \( (a + b)^n \) is given by: 

\(T_{k+1} = \binom{n}{k} a^{n-k} b^k\)

In this case, \( a = x \) and \( b = \frac{2}{x^3} \), and we are expanding \( \left(x + \frac{2}{x^3}\right)^{20} \). The general term in the expansion is:

\(T_{k+1} = \binom{20}{k} x^{20-k} \left(\frac{2}{x^3}\right)^k\)

We can simplify this term:

\(T_{k+1} = \binom{20}{k} x^{20-k} \cdot 2^k \cdot x^{-3k} = \binom{20}{k} 2^k x^{20 - k - 3k} = \binom{20}{k} 2^k x^{20 - 4k}\)

For the term to be independent of \( x \), the exponent of \( x \) must be 0. So, we set:

\(20 - 4k = 0\)

Solving for \( k \), we get:

\(k = 5\)

Therefore, the term independent of \( x \) corresponds to \( k = 5 \), and the term is:

\(T_{6} = \binom{20}{5} 2^5\)

The term independent of \( x \) is \( \binom{20}{5} 2^5 \).