Question:

For $2 \le \, r \, \le \, n,\binom{n}{r}+2 \binom{n}{r-1}+\binom{n+2}{r}$ is equal to

Updated On: June 02, 2025
  • $\binom{n+1}{r-1}$
  • $2 \binom{n+1}{r+1}$
  • $2 \binom{n+1}$
  • $2 \binom{n+1}{r}$
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The Correct Option is D

Solution and Explanation

$\binom{n}{r}+2 \binom{n}{r-1}+\binom{n}{r-2}=\bigg[\binom{n}{r}+\binom{n}{r-1}\bigg]$
+$\bigg[\binom{n}{n-r}+\binom{n}{n-r}\bigg]=\binom{n+1}{r} + \binom{n+1}{r-1}=\binom{n+1}{r}$
$[\because \, ^nC_r + \, ^nC_{r-1}= \, ^{n+1}C_r]$
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Concepts Used:

Binomial Theorem

The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is 

Properties of Binomial Theorem

  • The number of coefficients in the binomial expansion of (x + y)n is equal to (n + 1).
  • There are (n+1) terms in the expansion of (x+y)n.
  • The first and the last terms are xn and yn respectively.
  • From the beginning of the expansion, the powers of x, decrease from n up to 0, and the powers of a, increase from 0 up to n.
  • The binomial coefficients in the expansion are arranged in an array, which is called Pascal's triangle. This pattern developed is summed up by the binomial theorem formula.