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Question: \(nC_{0} +^{n + 1}C_{1} +^{n + 2}C_{2} + ... +^{n + 4}C_{r}\) is equal to...

nC0+n+1C1+n+2C2+...+n+4CrnC_{0} +^{n + 1}C_{1} +^{n + 2}C_{2} + ... +^{n + 4}C_{r} is equal to

A

n+rCrn + rC_{r}

B

n+r+1Crn + r + 1C_{r}

C

n+r+1Cr+1n + r + 1C_{r + 1}

D

None

Answer

n+r+1Crn + r + 1C_{r}

Explanation

Solution

Now,

nC0+n+1C1+n+2C2+...+n+rCrnC_{0} +^{n + 1}C_{1} +^{n + 2}C_{2} + ... +^{n + r}C_{r}

= (n+1C0+n+1C1)+n+2C2+n+3C3+....+n+rCr\left( n + 1C_{0} +^{n + 1}C_{1} \right) +^{n + 2}C_{2} +^{n + 3}C_{3} + .... +^{n + r}C_{r}

(∵ 6munC0=1 6mu= 6mun+1C0)\because\mspace{6mu}^{n}C_{0} = 1\mspace{6mu} = \mspace{6mu}^{n + 1}C_{0})

= (n+2C1+n+2C2)+n+3C3+....+n+rCr\left( n + 2C_{1} +^{n + 2}C_{2} \right) +^{n + 3}C_{3} + .... +^{n + r}C_{r}

(∵ 6mumCr+mCr−1=m+1Cr)\because\mspace{6mu}^{m}C_{r} +^{m}C_{r - 1} =^{m + 1}C_{r})

= n+3C2+n+3C3+...+n+rCrn + 3C_{2} +^{n + 3}C_{3} + ... +^{n + r}C_{r}

.............................

............................. continuing this way

= n+r+1Crn + r + 1C_{r}