Question

The middle term in the expansion of $\left(1+\frac{1}{5}\right)^{20}$ is

• A. ${}^{20}C_{9}(\frac{1}{5})^{9}$
• B. ${}^{20}C_{10}(\frac{1}{5})^{10}$
• C. ${}^{20}C_{11}(\frac{1}{5})^{11}$
• D. $(\frac{1}{5})^{11}$
• E. $(\frac{1}{5})^{10}$

Hint:

Binomial Theorem Tip: Remember that $r$ is always one less than the term number you are trying to find! For the 11th term, $r = 10$. This prevents common off-by-one errors.

Answer 1

The Correct Option is B

Concept:
In the binomial expansion of $(x+y)^n$, the total number of terms is always $n + 1$. The general term is given by the formula $T_{r+1} = {}^nC_r x^{n-r} y^r$. If $n$ is an even number, there is exactly one middle term located at the $(\frac{n}{2} + 1)$-th position.

Step 1: Determine the total number of terms.
The exponent in our binomial expansion is $n = 20$. The total number of terms in the expansion is $n + 1 = 20 + 1 = 21$. Since 21 is an odd number, there will be exactly one middle term.

Step 2: Find the position of the middle term.
The position of the single middle term for an even power $n$ is given by the formula $\frac{n}{2} + 1$: $$\text{Position} = \frac{20}{2} + 1 = 10 + 1 = 11$$ We need to find the 11th term, $T_{11}$.

Step 3: Set up the general term formula.
To find $T_{11}$ using the formula $T_{r+1} = {}^nC_r x^{n-r} y^r$, we must set $r = 10$ so that $r+1 = 11$. Identify the variables from the binomial $(1 + \frac{1}{5})^{20}$: $n = 20$, $x = 1$, $y = \frac{1}{5}$

Step 4: Substitute values into the general formula.
Plug the identified values into the term formula for $r=10$: $$T_{11} = {}^{20}C_{10} (1)^{20-10} \left(\frac{1}{5}\right)^{10}$$

Step 5: Simplify to match the options.
Since 1 raised to any power is simply 1, the $(1)^{10}$ term drops out of the equation entirely: $$T_{11} = {}^{20}C_{10} (1) \left(\frac{1}{5}\right)^{10}$$ $$T_{11} = {}^{20}C_{10} \left(\frac{1}{5}\right)^{10}$$ Hence the correct answer is (B) ${^{20}C_{10}(\frac{1}{5})^{10}$}.