Question

If \({}^{20}C_r\) is the co-efficient of \(x^r\) in the expansion of \((1+x)^{20\), then the value of \(\sum_{r=0}^{20} r^2 \cdot {}^{20}C_r\) is equal to :}

• A. \(420 \times 2^{19}\)
• B. \(380 \times 2^{19}\)
• C. \(380 \times 2^{18}\)
• D. \(420 \times 2^{18}\)

Hint:

Common binomial sums:
\(\sum r C_r = n 2^{n-1}\)
\(\sum r^2 C_r = n(n+1) 2^{n-2}\)

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a standard sum of binomial coefficients weighted by \(r^2\). It can be evaluated using differentiation of the binomial theorem expansion.
Step 2: Key Formula or Approach:
Use \(\sum_{r=0}^n r^2 \binom{n}{r} = n(n+1)2^{n-2}\).
Step 3: Detailed Explanation:
Start with \((1+x)^n = \sum_{r=0}^n \binom{n}{r} x^r\).
Differentiate with respect to \(x\):
\[ n(1+x)^{n-1} = \sum_{r=0}^n r \binom{n}{r} x^{r-1} \]
Multiply by \(x\):
\[ nx(1+x)^{n-1} = \sum_{r=0}^n r \binom{n}{r} x^r \]
Differentiate again:
\[ n(1+x)^{n-1} + nx(n-1)(1+x)^{n-2} = \sum_{r=0}^n r^2 \binom{n}{r} x^{r-1} \]
Substitute \(x = 1\):
\[ \sum_{r=0}^n r^2 \binom{n}{r} = n \cdot 2^{n-1} + n(n-1)2^{n-2} = 2^{n-2}(2n + n^2 - n) = n(n+1)2^{n-2} \]
For \(n = 20\):
\[ \text{Sum} = 20(21) \cdot 2^{20-2} = 420 \cdot 2^{18} \]
Step 4: Final Answer:
The sum is equal to \(420 \times 2^{18}\).