Question

The greatest term in the expansion of $ (1+x)^{15} $, when $ x = \frac{1}{2} $ is

• A. $ \frac{1}{32} {}^{15}C_5 $
• B. $ \frac{1}{64} {}^{15}C_6 $
• C. $ \frac{1}{32} {}^{15}C_6 $
• D. $ \frac{1}{64} {}^{15}C_5 $

Hint:

Find $ r $ such that $ \frac{T_{r+1}}{T_r} \ge 1 $ and $ \frac{T_r}{T_{r-1}} \ge 1 $. The greatest term is $ T_{r+1} $.

Answer 1

The Correct Option is A

Step 1: Write the general term in the binomial expansion of $ (1+x)^n $.
$ T_{r+1} = {}^{n}C_r x^r $.
Step 2: Apply the given values of $ n $ and $ x $.
$ n = 15 $, $ x = \frac{1}{2} $.
$ T_{r+1} = {}^{15}C_r \left(\frac{1}{2}\right)^r = \frac{{}^{15}C_r}{2^r} $.
Step 3: Find the ratio of consecutive terms $ \frac{T_{r+1}{T_r} $.}
$ \frac{T_{r+1}}{T_r} = \frac{{}^{15}C_r / 2^r}{{}^{15}C_{r-1} / 2^{r-1}} = \frac{{}^{15}C_r}{2 {}^{15}C_{r-1}} $. 
Step 4: Use the formula for the ratio of binomial coefficients.
$ \frac{{}^{n}C_r}{{}^{n}C_{r-1}} = \frac{n - r + 1}{r} $.
$ \frac{{}^{15}C_r}{{}^{15}C_{r-1}} = \frac{16 - r}{r} $.
Step 5: Substitute this ratio back into the expression for $ \frac{T_{r+1}{T_r} $.}
$ \frac{T_{r+1}}{T_r} = \frac{1}{2} \cdot \frac{16 - r}{r} = \frac{16 - r}{2r} $. 
Step 6: Find the value of $ r $ for which the term is the greatest.
$ \frac{T_{r+1}}{T_r} \ge 1 \implies \frac{16 - r}{2r} \ge 1 \implies 16 - r \ge 2r \implies 16 \ge 3r \implies r \le \frac{16}{3} = 5.33 $.
The largest integer $ r $ is 5.
Step 7: Determine the greatest term.
The greatest term is $ T_{5+1} = T_6 = \frac{{}^{15}C_5}{2^5} = \frac{{}^{15}C_5}{32} $. 
Step 8: Conclusion.
The greatest term is $ \frac{1}{32} {}^{15}C_5 $.