If the coefficient of `r^(th)` term, `(r+4)^(th)` term are equal in the expansion of `(1 + x)^(20)`, then the value of r will be

Options

A. 6
B. 8
C. 9
D. 11

Answer

To solve the problem, we need to find the value of \( r \) such that the coefficients of the \( r^{th} \) term and the \( (r+4)^{th} \) term in the expansion of \( (1 + x)^{20} \) are equal.

### Step 1: Identify the coefficients of the terms
In the binomial expansion of \( (1 + x)^{20} \), the \( r^{th} \) term is given by:
\[ T_r = \binom{20}{r-1} x^{r-1} \]
The coefficient of the \( r^{th} \) term is \( \binom{20}{r-1} \).

The \( (r+4)^{th} \) term is given by:
\[ T_{r+4} = \binom{20}{(r+4)-1} x^{(r+4)-1} = \binom{20}{r+3} x^{r+3} \]
The coefficient of the \( (r+4)^{th} \) term is \( \binom{20}{r+3} \).

### Step 2: Set the coefficients equal
We set the coefficients of the two terms equal to each other:
\[ \binom{20}{r-1} = \binom{20}{r+3} \]

### Step 3: Apply the property of binomial coefficients
Using the property of binomial coefficients, we know that:
\[ \binom{n}{k} = \binom{n}{n-k} \]
Thus, we can write:
\[ \binom{20}{r+3} = \binom{20}{20 - (r+3)} = \binom{20}{17 - r} \]
So, we have:
\[ \binom{20}{r-1} = \binom{20}{17 - r} \]

### Step 4: Set the lower indices equal
Since the binomial coefficients are equal, we can equate their lower indices:
\[ r - 1 = 17 - r \]

### Step 5: Solve for \( r \)
Now, we solve for \( r \):
\[ r - 1 + r = 17 \]
\[ 2r - 1 = 17 \]
\[ 2r = 18 \]
\[ r = 9 \]

### Conclusion
The value of \( r \) is \( 9 \).

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