Binomial Theorem Applications

Question

The value of C(n,0) + C(n,1) + C(n,2) + ... + C(n,n) equals

RankGuru · Accepted Answer

1. **Identity Proof**: Start with the standard expansion:
   - $(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n$
2. **Substitution**: Let $x = 1$.
   - $(1+1)^n = \binom{n}{0} + \binom{n}{1}(1) + \binom{n}{2}(1)^2 + \dots + \binom{n}{n}(1)^n$
3. **Result**: $2^n = \sum_{r=0}^{n} \binom{n}{r}$. The sum of all binomial coefficients for a power $n$ is $2^n$.

Rank Guru

Binomial Theorem Applications

Select the correct option:

Solution

🔒 Solution Hidden from View

More binomial theorem applications Practice Questions

Using binomial theorem, (99)⁵ can be expressed as

The sum C(n,0) - C(n,1) + C(n,2) - C(n,3) + ... equals

The remainder when 2¹⁰⁰ is divided by 7 is

About This Question

Solution