Question

The coefficients of three consecutive terms of $(1+x{)}^{n+5}$ are in the ratio $5:10:14$. Then $n=$ ____ .

Short answer

n = 6

Step-by-step solution

Understand the Binomial Theorem

The Binomial Theorem states that the expansion of (1+x)^n has coefficients determined by the binomial coefficients which are given by the formula C(n, k) = n! / [k!(n-k)!] where C(n, k) is the coefficient of the k-th term, x^k.

Set Up the Ratio of Consecutive Terms

The ratio of the coefficients of the three consecutive terms is 5:10:14. Let the three consecutive terms have binomial coefficients C(n+5, k), C(n+5, k+1), and C(n+5, k+2) respectively. Thus, we have the following ratios: C(n+5, k) / C(n+5, k+1) = 5/10 = 1/2 and C(n+5, k+1) / C(n+5, k+2) = 10/14 = 5/7.

Apply the Binomial Coefficient Ratio Formula

For binomial coefficients, C(n, k) / C(n, k+1) = (n-k) / (k+1). Apply this formula to the two ratios we have obtained to get two equations: (n+5-k) / (k+1) = 1/2 and (n+5-(k+1)) / (k+2) = 5/7.

Solve the Equations Simultaneously

We have two equations with two variables (n and k) that we can solve simultaneously. From the first equation, we get 2(n+5-k) = (k+1), which simplifies to 2n + 10 - 2k = k + 1. Clean it up to get the first relationship between n and k: 2n - 3k = -9. From the second equation, we get 7(n+4-k) = 5(k+2), which simplifies to 7n + 28 - 7k = 5k + 10. Rearranging the terms gives us the second relationship: 7n - 12k = -18.

Find the value of n

Now we have a system of linear equations: 2n - 3k = -9 and 7n - 12k = -18. We can solve for n by multiplying the first equation by 7 and the second by 2, and then subtracting the two resulting equations to eliminate k. Upon solving, we find that n = 6.

Key concepts

Binomial coefficients

Binomial coefficients are fundamental when working with the Binomial Theorem. In simplified terms, they tell us the number of ways to choose a subset of elements from a larger set, and this concept is crucial in combinatorics. For example, when expanding $(1+x{)}^n$ using the Binomial Theorem, the coefficient of the k-th term, usually represented as $C(n,k)$, is calculated using the formula $C(n,k)=\frac{n!}{k!(n-k)!}$. This expresses the number of combinations of n items taken k at a time, where $!$ denotes factorial.

Understanding binomial coefficients is an important step in finding the relationship between the terms of a binomial expansion. These coefficients underpin many principles of probability and statistics, making them a cornerstone concept for students prepping for competitive exams like the JEE Advanced.

Binomial expansion

The Binomial expansion is the process of expanding an expression raised to any power, described by the Binomial Theorem. Specifically, for the expression $(1+x{)}^n$ where n is a non-negative integer, the expansion is a sum of terms in the form $C(n,k)x^k$, with the binomial coefficients $C(n,k)$ determining the weight of each term. The Binomial Theorem is beautifully symmetric and helps in understanding patterns of polynomial expansions.

An important aspect of these expansions is that the exponents of $x$ in the terms decrease by one in each subsequent term, starting from $n$ down to zero, while the binomial coefficients correspondingly change, following their combinatorial definition. In our exercise, recognizing this allows us to set up and solve for the unknown power in the binomial expansion by exploring the relationship of the provided coefficients.

Ratio of consecutive terms

In the context of binomial expansions, the ratio of consecutive terms can provide valuable insights about the underlying algebraic structure. By examining the ratio of the coefficients of consecutive terms, we can infer properties about the expansion and solve for variables of interest such as $n$ in our given exercise.

The ratio between binomial coefficients $C(n,k)$ and $C(n,k+1)$ simplifies to $\frac{n-k}{k+1}$ due to the properties of factorial in their formula. This simplification facilitates the derivation of a proportional relationship that can then be used to set up equations, as we observed in the solution for the exercise. Understanding how the ratios work is critical especially in problems often encountered in JEE Advanced exams, where the manipulation of such properties can lead to a swift and elegant solution.