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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Binomial Theorem

NDA 2018 · Mathematics · Binomial Theorem

What is C(n, r)+2 C(n, r-1)+C(n, r-2) equal to?

What is C(n,r)+2C(n,r−1)+C(n,r−2)C(n, r)+2 C(n, r-1)+C(n, r-2) equal to?

Options

  1. AC(n+1,r)\mathrm{C}(\mathrm{n}+1, \mathrm{r})
  2. BC(n−1,r+1)\mathrm{C}(\mathrm{n}-1, \mathrm{r}+1)
  3. CC(n,r+1)C(n, r+1)
  4. DC(n+2,r)\mathrm{C}(\mathrm{n}+2, \mathrm{r})

Correct answer

D. C(n+2,r)\mathrm{C}(\mathrm{n}+2, \mathrm{r})

Step-by-step solution

C(n,r)+2C(n,r−1)+C(n,r−2)=nCr+2(nCr−1)+nCr−2=nCr+nCr−1+nCr−1+nCr−2(∵nCr+nCr−1=n+1Cr)\begin{aligned} & C(n, r)+2 C(n, r-1)+C(n, r-2) \\ & ={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}+2\left({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}\right)+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-2} \\ & ={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-2} \\ & \left(\because{ }^{\mathrm{n}} C_r+{ }^n C_{r-1}={ }^{n+1} C_r\right) \end{aligned}=n+1Cr+(n+1r−1)=n+2Cr=C(n+2,r)\begin{aligned} & ={ }^{n+1} C_r+\binom{n+1}{r-1} \\ & ={ }^{n+2} C_r=C(n+2, r)\end{aligned}$
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