KCET · Maths · Binomial Theorem
In the expansion of \((1+x)^n\)
\(\frac{C_1}{C_0}+2 \frac{C_2}{C_1}+3 \frac{C_3}{2}+\ldots+n \frac{C_n}{C_{n-1}}\) is equal to
- A \(\frac{n(n+1)}{2}\)
- B \(\frac{n}{2}\)
- C \(\frac{n+1}{2}\)
- D \(3 n(n+1)\)
Answer & Solution
Correct Answer
(A) \(\frac{n(n+1)}{2}\)
Step-by-step Solution
Detailed explanation
\(\because \frac{C_1}{C_0}+2 \cdot \frac{C_2}{C_1}+3 \cdot \frac{C_3}{C_2}+\ldots n \frac{C_n}{C_{n-1}}\)
\(=\frac{n}{1}+2 \cdot \frac{\frac{n(n-1)}{1 \cdot 2}}{n}+3 \cdot \frac{\frac{n(n-1)(n-2)}{3 \cdot 2 \cdot 1}}{\frac{n(n-1)}{n}}+\ldots+n \cdot \frac{1}{n}\)
\(=n+(n-1)+(n-2) \ldots+1=\Sigma n=\frac{n(n+1)}{2}\)
\(=\frac{n}{1}+2 \cdot \frac{\frac{n(n-1)}{1 \cdot 2}}{n}+3 \cdot \frac{\frac{n(n-1)(n-2)}{3 \cdot 2 \cdot 1}}{\frac{n(n-1)}{n}}+\ldots+n \cdot \frac{1}{n}\)
\(=n+(n-1)+(n-2) \ldots+1=\Sigma n=\frac{n(n+1)}{2}\)
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