Question

The \(11^{\text {th }}\) term from last, in expansion of \(\left[2 \mathrm{x}+\left(1 / \mathrm{x}^{2}\right)\right]^{25}\) is (a) \(-{ }^{25} \mathrm{C}_{15}\left(2^{10} / \mathrm{x}^{20}\right)\) (b) \(+{ }^{25} \mathrm{C}_{15}\left(2^{10} / \mathrm{x}^{20}\right]\) (c) \(-{ }^{25} \mathrm{C}_{14}\left(2^{11} / \mathrm{x}^{11}\right)\) (d) \({ }^{25} \mathrm{C}_{14}\left(2^{11} / \mathrm{x}^{11}\right)\)

Short answer

The short answer is: \({ }^{25}C_{14}\ 2^{11}x^{-11}\)

Step-by-step solution

Using binomial theorem to find general term

We can expand the given expression using the binomial theorem. For any positive integer n, \((a+b)^n\) can be expanded as \[\sum_{k=0}^{n}\ ^nC_k\ a^{(n-k)}\ b^k\] where \(_{n} C_{k}\) represents the binomial coefficients. Now apply this to the given expression \(\left[2x+\left(\frac{1}{x^2}\right)\right]^{25}\): Then, the general term in the expansion will be \[^{25}C_k\ (2x)^{(25-k)}\left(\frac{1}{x^2}\right)^k\]

Simplifying the general term

Next, we will rewrite and simplify the general term based on the binomial theorem: \[^{25}C_k\ (2^{25-k}x^{(25-k)})\left(\frac{1}{x^{2k}}\right)\] which simplifies to \[^{25}C_k\ 2^{25-k}x^{25-3k}\]

Finding the 11th term from the last

To find the 11th term from the last, we need to find the corresponding term from the beginning. Here, the total number of terms in the expansion is 26 (from \(k=0\) to \(k=25\), inclusive), so the 11th term from the last is equivalent to the 16th term from the beginning (26 - 11 = 15, means k=14). Hence, substitute \(k=14\) in the simplified general term: \[^{25}C_{14}\ 2^{25-14}\ x^{25-3 \times 14} = { }^{25}C_{14}\ 2^{11}x^{-11}\]

Comparing and choosing the correct option

We now have the 11th term from the last in the expanded form: \[^{25}C_{14}\ 2^{11}x^{-11}\] Compare this with the given options to identify the correct one: (a) \(-{ }^{25}C_{15}\left(2^{10}\ /x^{20}\right)\): incorrect (b) \(+{ }^{25}C_{15}\left(2^{10}\ /x^{20}\right)\): incorrect (c) \(-{ }^{25}C_{14}\left(2^{11}\ /x^{11}\right)\): incorrect (d) \(+{ }^{25}C_{14}\left(2^{11}\ /x^{11}\right)\): correct The correct answer is (d) \(+{ }^{25}C_{14}\left(2^{11}\ /x^{11}\right)\).

Key concepts

Binomial Expansion

In mathematics, the binomial expansion is a method for expressing any power of a binomial, which is an expression consisting of two terms, as a sum involving terms of the form \(a^{m}b^{n}\). A binomial expression might look like \((a + b)^n\), and binomial expansion provides a formula for expanding this expression:

\[(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]This formula not only helps expand powers of binomials but also illustrates how each term in the expansion is formed by coefficients given by binomial coefficients \(\binom{n}{k}\) and powers of each term in the binomial.It proves very useful in algebra and calculus for simplifying expressions, and it forms the foundation for understanding multinomial expansions and higher algebra concepts. In this specific exercise, we are using binomial expansion to break down\[ (2x + \frac{1}{x^2})^{25} \]into a series of terms.

Binomial Coefficients

Binomial coefficients are essential components of the binomial theorem. They denote the coefficients in the binomial expansion and are represented by \(\binom{n}{k}\), which reads as "n choose k." Binomial coefficients count the number of ways to choose \(k\) elements from a set of \(n\) elements, without considering the order of selection. Mathematically, binomial coefficients are calculated using the formula:

\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]where \(n!\) is the factorial of \(n\). This means multiplying all positive integers up to and including \(n\).In the context of the exercise, binomial coefficients determine the weight or magnitude of each term in the expansion of \((2x + \frac{1}{x^2})^{25}\). Specifically, \(\binom{25}{14}\) is used for finding a particular term from the expansion, which indicates how many times a particular arrangement appears at the \(14^{th}\) term from the start, or equivalently in this problem, the \(11^{th}\) term from the end.

General Term in Binomial Expansion

The general term in a binomial expansion is a critical element as it allows us to find any specific term within the expanded expression.It is represented generally by the formula:

\[T_{k+1} = \binom{n}{k} a^{n-k} b^k\]This formula provides the \((k+1)\)th term in the expansion of \((a+b)^n\), which is especially useful for identifying and calculating any term without expanding the entire expression. In the exercise at hand, we make use of this to find the needed term from the expansion of \((2x + \frac{1}{x^2})^{25}\). By plugging in specific values of \(k\) (for example, \(k=14\) for the \(11^{th}\) term from last), we can directly find and simplify the term using this formula. This methodology helps effortlessly solve problems related to the binomial expansion and is a powerful tool in combinatorial mathematics.