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In the expansion of (1 + x)ⁿ , $\frac{C_{1}}{C_{0}} + 2 \frac{C_{2}}{C_{1}} + 3 \frac{C_{3}}{C_{2}} + \dots + n \frac{C_{n}}{C_{n - 1}}$ is equal to

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(n+1)2

n(n+1)2

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detailed solution

Correct option is C

Since, we know that nCr nCr−1=n−r+1r∴C1C0=n,C2C1=n−12,C3C2=n−23,… Thus, C1C0+2C2C1+3C3C2+…=n+2⋅n−12+3⋅n−23+…+n⋅1n=[n+(n−1)+(n−2)+…+1]=n(n+1)2

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In the expansion of (1 + x)n ,C1C0+2C2C1+3C3C2+…+nCnCn−1 is equal to

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In the expansion of (1 + x)ⁿ , $\frac{C_{1}}{C_{0}} + 2 \frac{C_{2}}{C_{1}} + 3 \frac{C_{3}}{C_{2}} + \dots + n \frac{C_{n}}{C_{n - 1}}$ is equal to