© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the problem
We want to prove that
$^n C_{r+1} + \; ^n C_{r-1} + 2 \times \, ^n C_r = \, ^{n+2} C_{r+1}.$
Recall that $^n C_r$ (also written as $C(n,r)$) represents the number of ways to choose $r$ items from $n$ items.
Step 2: Recall an important combination identity
The standard identity we will use is:
$^n C_r \;+\; ^n C_{r-1} \;=\; ^{n+1} C_r.$
This identity allows us to merge or split combination terms where indices differ by 1.
Step 3: Group the given expression
Consider the expression:
$^n C_{r+1} + 2 \times \, ^n C_r + \, ^n C_{r-1}.$
We rewrite the middle term and combine them in pairs:
$^n C_{r+1} + ^n C_r$
$^n C_r + ^n C_{r-1}$
This gives us:
$\bigl(^n C_{r+1} + ^n C_r\bigr) + \bigl(^n C_r + ^n C_{r-1}\bigr).$
Step 4: Apply the combination identity to each pair
From the identity $^n C_r + \, ^n C_{r-1} = \, ^{n+1} C_r$, we get:
$^n C_{r+1} + ^n C_r = \, ^{n+1} C_{r+1}.$
$^n C_r + ^n C_{r-1} = \, ^{n+1} C_r.$
Hence, the original expression becomes:
$^{n+1} C_{r+1} + \, ^{n+1} C_r.$
Step 5: Simplify to get the final result
We now use the same identity again, but with $n$ replaced by $(n+1)$ and $r$ replaced by $(r+1 - 1)=r$:
$^{n+1} C_{r+1} + \, ^{n+1} C_r = \, ^{n+2} C_{r+1}.$
Therefore, the expression indeed simplifies as:
$^n C_{r+1} + 2 \times \, ^n C_r + ^n C_{r-1} = \, ^{n+2} C_{r+1}.$
Conclusion
The correct answer is $^{n+2}C_{r+1}$, verifying that
$^n C_{r+1} + 2 \times \, ^n C_r + \, ^n C_{r-1} = \, ^{n+2} C_{r+1}.$
