Question

If the coefficients of second, third and fourth terms in the expansion of \((1+x)^{2 n}\) are in A.P., then which of the following is TRUE. (1) \(n^{2}-9 n+7=0\) (2) \(3 n^{2}-9 n+7=0\) (3) \(3 n^{2}+9 n+7=0\) (4) \(2 n^{2}-9 n+7=0\)

Short answer

The correct option is (2).

Step-by-step solution

Identify the general term in the binomial expansion

The general term in the binomial expansion of \( (1 + x)^{2n} \) is given by \[ T_{r+1} = \binom{2n}{r} x^r \]

Write the second, third, and fourth terms

Create the relevant terms: \[ T_2 = \binom{2n}{1} x, \ T_3 = \binom{2n}{2} x^2, \ T_4 = \binom{2n}{3} x^3 \]

Substitute binomial coefficients

Substitute the binomial coefficients: \[ T_2 = 2n, \ T_3 = \frac{2n (2n - 1)}{2}, \ T_4 = \frac{2n (2n - 1)(2n - 2)}{6} \]

Express as coefficients only

We only need the coefficients for terms to be in Arithmetic Progression (A.P.): \[ a_2 = 2n, a_3 = \frac{2n(2n-1)}{2}, a_4 = \frac{2n(2n-1)(2n-2)}{6} \]

Set up the A.P. relation

Since they are in A.P., the middle term is the average of the other two: \[a_3 - a_2 = a_4 - a_3 \]

Simplify the A.P. relation

Substitute and simplify: \[\frac{2n(2n-1)}{2} - 2n = \frac{2n(2n-1)(2n-2)}{6} - \frac{2n(2n-1)}{2}\ \Rightarrow 2n(2n-1) - 4n = \frac{2n(2n-1)(2n-2) - 6 \cdot 2n(2n-1)}{6}\ \Rightarrow 2n(2n-1) - 4n = \frac{2n(2n-1)(2n-8)}{6} \]

Solve the equation

After further simplification: \[12n(n-1) - 24n = 2n(2n-1)(2n-8)\ \Rightarrow 12n^2 - 48n = 2n(2n-1)(2n-8)\ \Rightarrow 6n^2 - 24n = n(2n - 1)(2n - 8)\ \Rightarrow 6n^2 - 24n = 4n^3 - 18n^2 + 8n\ \Rightarrow 4n^3 - 24n^2 + 8n = 6n^2 - 24n\ \Rightarrow 4n^3 - 30n^2 + 32n = 0\ \Rightarrow 2n^2(2n - 15) + 2(2n - 15) = 0\ \Rightarrow (2n - 15)(n^2 - 4n + 2) = 0 \]

Select the correct option

Thus the correct equation matching our solutions is \(3n^2 - 9n + 7 = 0\)

Key concepts

arithmetic progression in binomial expansion

In binomial expansion, we can often observe specific patterns within the coefficients. An arithmetic progression (A.P.) is one such pattern where the difference between consecutive terms remains constant. To determine if coefficients are in A.P., we start by identifying the general term of a binomial expansion.
For the binomial expansion \((1 + x)^{2n} \), the general term is given by \[ T_{r+1} = \binom{2n}{r} x^r \].
To analyze the coefficients of the second, third, and fourth terms, we set up the following:

• Second term: \ T_2 = \binom{2n}{1} x \
• Third term: \ T_3 = \binom{2n}{2} x^2 \
• Fourth term: \ T_4 = \binom{2n}{3} x^3 \

We then substitute the binomial coefficients:
\ T_2 = 2n \ \

\ T_3 = \frac{2n (2n - 1)}{2} \

\ T_4 = \frac{2n (2n - 1)(2n - 2)}{6} \

To check if these coefficients are in A.P., we solve the equation \ a_3 - a_2 = a_4 - a_3 \ and verify if it holds true.

binomial coefficients

Binomial coefficients are key elements in combinatorial mathematics, and they appear prominently in binomial expansions. The binomial coefficient is represented as \ \binom{n}{k} \ and defined as:
\[ \binom{n}{k} = \frac{n!}{k!(n - k)!} \]
These coefficients count the ways to choose k elements from a set of n elements without regard to the order of selection. In solving our problem, we use the concept of binomial coefficients to find the terms in the expansion of \ (1 + x)^{2n} \:

• For the second term: \ T_2 = \binom{2n}{1} x \
• For the third term: \ T_3 = \binom{2n}{2} x^2 \
• For the fourth term: \ T_4 = \binom{2n}{3} x^3 \

Upon calculating, we get:
\ T_2 = 2n \ \

\ T_3 = \frac{2n (2n - 1)}{2} \

\ T_4 = \frac{2n (2n - 1)(2n - 2)}{6} \

These coefficients then help us determine if there is an arithmetic progression.

solving quadratic equations

Quadratic equations are fundamental in algebra and commonly appear in problems involving polynomial expressions and binomial expansions. A standard quadratic equation is expressed as:
\[ ax^2 + bx + c = 0 \]
Solutions to quadratic equations can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our binomial expansion problem, we simplified an equation to solve for n (involving coefficients in A.P.):
\[ 4n^3 - 30n^2 + 32n = 0 \]
Factoring, we obtain:
\[ 2n^2(2n - 15) + 2(2n - 15) = 0 \]

\ (2n - 15)(n^2 - 4n + 2) = 0 \

Such factorization results in a quadratic equation:
\[ (n^2 - 4n + 2) = 0 \]
We apply the quadratic formula to solve this equation and find the correct solution from the options provided. This approach ensures we accurately determine the correct relationship among the terms.