What is C(n, 1) + C(n, 2) + _ _ _ _ _ + C(n, n) equal to
1 + 2 + 2 2 + 2 3 + _ _ _ _ _ _ + 2 n - 1
Understanding the Sum of Binomial Coefficients C(n, k)
The question asks for the value of the sum of binomial coefficients from C(n, 1) up to C(n, n). This sum can be written as:
\(\sum_{k=1}^{n} C(n, k) = C(n, 1) + C(n, 2) + \dots + C(n, n)\)
Relating to the Binomial Theorem
The binomial theorem provides a way to expand \((a+b)^n\). For our purpose, let's consider the expansion of \((1+x)^n\):
\((1+x)^n = C(n, 0)x^0 + C(n, 1)x^1 + C(n, 2)x^2 + \dots + C(n, n)x^n\)
This can be written using summation notation as:
\((1+x)^n = \sum_{k=0}^{n} C(n, k) x^k\)
Using a Specific Case (x=1)
Let's substitute \(x=1\) into the binomial expansion:
\((1+1)^n = \sum_{k=0}^{n} C(n, k) 1^k\)
\(2^n = \sum_{k=0}^{n} C(n, k) \cdot 1\)
\(2^n = C(n, 0) + C(n, 1) + C(n, 2) + \dots + C(n, n)\)
This shows that the sum of all binomial coefficients for a given n, from k=0 to n, is equal to \(2^n\).
Finding the Required Sum
The sum we need to find is \(C(n, 1) + C(n, 2) + \dots + C(n, n)\). Notice that the total sum \(\sum_{k=0}^{n} C(n, k)\) includes the term \(C(n, 0)\), which is not in the required sum.
We know that:
\(\sum_{k=0}^{n} C(n, k) = C(n, 0) + [C(n, 1) + C(n, 2) + \dots + C(n, n)]\)
We also know that \(C(n, 0)\) is the number of ways to choose 0 items from n, which is always 1.
\(C(n, 0) = \frac{n!}{0!(n-0)!} = \frac{n!}{1 \cdot n!} = 1\)
So, substituting the values we have:
\(2^n = 1 + [C(n, 1) + C(n, 2) + \dots + C(n, n)]\)
Rearranging the equation to find the required sum:
\(C(n, 1) + C(n, 2) + \dots + C(n, n) = 2^n - 1\)
Checking the Options
Now let's examine the given options to see which one equals \(2^n - 1\). The options are presented as sums of powers of 2. These are geometric series.
The sum of a geometric series \(a + ar + ar^2 + \dots + ar^{m-1}\) with first term \(a\), common ratio \(r\), and \(m\) terms is given by \(S_m = a \frac{(r^m - 1)}{r - 1}\) (for \(r \neq 1\)).
Let's evaluate each option:
| Option | Series | First Term (a) | Ratio (r) | Number of Terms (m) | Sum (\(S_m\)) |
|---|---|---|---|---|---|
| 1 | \(2 + 2^2 + \dots + 2^n\) | 2 | 2 | n | \(2 \frac{(2^n - 1)}{2 - 1} = 2(2^n - 1) = 2^{n+1} - 2\) |
| 2 | \(1 + 2 + 2^2 + \dots + 2^n\) | 1 | 2 | n+1 | \(1 \frac{(2^{n+1} - 1)}{2 - 1} = 2^{n+1} - 1\) |
| 3 | \(1 + 2 + 2^2 + \dots + 2^{n-1}\) | 1 | 2 | n | \(1 \frac{(2^n - 1)}{2 - 1} = 2^n - 1\) |
| 4 | \(2 + 2^2 + \dots + 2^{n-1}\) | 2 | 2 | n-1 | \(2 \frac{(2^{n-1} - 1)}{2 - 1} = 2(2^{n-1} - 1) = 2^n - 2\) |
Comparing our result \(2^n - 1\) with the sums of the options, we see that Option 3 matches the required value.
Therefore, \(C(n, 1) + C(n, 2) + \dots + C(n, n)\) is equal to \(1 + 2 + 2^2 + \dots + 2^{n-1}\).
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