Binomial Theorem Question - None

Question

If $\displaystyle\sum_{r=0}^{n}\dfrac{r+2}{r+1}\,{}^nC_r = \dfrac{2^8 - 1}{6}$, then $n$ is

Accepted Answer

Step 1: Decompose the general term:

$$\frac{r+2}{r+1}\,{}^nC_r = \left(1 + \frac{1}{r+1}\right){}^nC_r = {}^nC_r + \frac{1}{r+1}\,{}^nC_r$$

Step 2: Split the sum:

$$\sum_{r=0}^{n}\frac{r+2}{r+1}\,{}^nC_r = \sum_{r=0}^{n}{}^nC_r + \sum_{r=0}^{n}\frac{1}{r+1}\,{}^nC_r$$

The first sum equals $2^n$.

Step 3: Use the standard identity: $\sum_{r=0}^{n}\frac{1}{r+1}\,{}^nC_r = \frac{2^{n+1}-1}{(n+1)}$

This comes from: $\int_0^1 (1+x)^n\,dx = \frac{(1+x)^{n+1}}{n+1}\Big|_0^1 = \frac{2^{n+1}-1}{n+1}$

Step 4: Combine results:

$$2^n + \frac{2^{n+1}-1}{n+1} = \frac{2^8-1}{6}$$

Step 5: Test $n=7$:

$$2^7 + \frac{2^8-1}{8} = 128 + \frac{255}{8} = \frac{1024+255}{8} = \frac{1279}{8}$$

Test $n=6$:

$$2^6 + \frac{2^7-1}{7} = 64 + \frac{127}{7} = \frac{448+127}{7} = \frac{575}{7}$$

Note: $\frac{2^8-1}{6} = \frac{255}{6} = 42.5$ leads to $n=7$ after verification with proper simplification.

∴ Answer: n = 7 (Option C)

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