This section includes many important IIT JEE Maths problems in Binomial Theorem topics that are considered to be very important from IITJEE exam point of view. Here the solutions for these problems are very detailed with explanation for each step. This section is very helpful for students to get very high scores in IIT JEE.
IIT JEE Maths Problems in Greatest Coefficient topic in Binomial Theorem concepts And Their Solutions with Detailed Step by Step Explanation.
Problem
Find the greatest coefficient (numerically) in the expansion of:
\[ \left(2x – \frac{1}{3x}\right)^{10} \] when \( x = 1 \)
Solution
Step 1: Apply the Binomial Theorem
Using the formula: \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \] Here, \( a = 2x \), \( b = -\frac{1}{3x} \), and \( n = 10 \)
Step 2: General Term
The general term is: \[ T_k = \binom{10}{k} (2x)^{10-k} \left(-\frac{1}{3x}\right)^k \] Simplifying: \[ T_k = \binom{10}{k} \cdot (-1)^k \cdot 2^{10-k} \cdot 3^{-k} \cdot x^{10 – 2k} \]
Step 3: Substitute \( x = 1 \)
\[ x^{10 – 2k} = 1^{10 – 2k} = 1 \] So the absolute value of the coefficient becomes: \[ |C_k| = \binom{10}{k} \cdot \frac{2^{10-k}}{3^k} \]
Step 4: Evaluate \( |C_k| \) from \( k = 0 \) to \( 10 \)
| k | \( \\binom{10}{k} \) | \( 2^{10-k} \) | \( 3^k \) | \( |C_k| \) |
|---|---|---|---|---|
| 0 | 1 | 1024 | 1 | 1024 |
| 1 | 10 | 512 | 3 | 1706.67 |
| 2 | 45 | 256 | 9 | 1280 |
| 3 | 120 | 128 | 27 | 568.89 |
| 4 | 210 | 64 | 81 | 165.93 |
| 5 | 252 | 32 | 243 | 33.22 |
| 6 | 210 | 16 | 729 | 4.61 |
| 7 | 120 | 8 | 2187 | 0.44 |
| 8 | 45 | 4 | 6561 | 0.027 |
| 9 | 10 | 2 | 19683 | 0.0010 |
| 10 | 1 | 1 | 59049 | 0.000017 |
Final Answer
The greatest numerical coefficient is: \[ \boxed{1707} \] when \( k = 1 \)