Let's consider the polynomial expansion of [tex]\((1 + x)^n\)[/tex]:
[tex]\[
(1 + x)^n = C_0 + C_1 x + C_2 x^2 + \cdots + C_n x^n
\][/tex]
Here, [tex]\(C_k\)[/tex] are the binomial coefficients given by:
[tex]\[
C_k = \binom{n}{k}
\][/tex]
We are required to prove that:
[tex]\[
C_1 + 2C_2 + 3C_3 + \cdots + nC_n - \frac{1}{2}(n \cdot 2^n) = 0
\][/tex]
First, recall what the binomial theorem tells us:
[tex]\[
(1 + x)^n = \sum_{k=0}^n \binom{n}{k} x^k
\][/tex]
### Step 1: Differentiate Both Sides
To find [tex]\(C_1 + 2C_2 + 3C_3 + \cdots + nC_n\)[/tex], we differentiate [tex]\((1 + x)^n\)[/tex] with respect to [tex]\(x\)[/tex]:
[tex]\[
\frac{d}{dx} \left((1 + x)^n\right) = \frac{d}{dx} \left(\sum_{k=0}^n \binom{n}{k} x^k\right)
\][/tex]
The left-hand side, by using the chain rule, gives:
[tex]\[
\frac{d}{dx} \left((1 + x)^n\right) = n(1 + x)^{n-1}
\][/tex]
The right-hand side, using term-by-term differentiation, gives:
[tex]\[
\sum_{k=0}^n \binom{n}{k} \frac{d}{dx} (x^k) = \sum_{k=1}^n k \binom{n}{k} x^{k-1}
\][/tex]
Note that the [tex]\(k = 0\)[/tex] term vanishes since the derivative of [tex]\(x^0\)[/tex] is 0.
### Step 2: Evaluate at [tex]\(x = 1\)[/tex]
Now, set [tex]\(x = 1\)[/tex] in both expressions. For the left-hand side:
[tex]\[
n(1 + 1)^{n-1} = n \cdot 2^{n-1}
\][/tex]
For the right-hand side:
[tex]\[
\sum_{k=1}^n k \binom{n}{k} \cdot 1^{k-1} = \sum_{k=1}^n k \binom{n}{k}
\][/tex]
Therefore, equating these, we get:
[tex]\[
\sum_{k=1}^n k \binom{n}{k} = n \cdot 2^{n-1}
\][/tex]
### Step 3: Simplify the Desired Expression
The expression we want to prove can be rewritten using [tex]\(\sum_{k=1}^n k \binom{n}{k}\)[/tex]:
[tex]\[
C_1 + 2C_2 + 3C_3 + \cdots + nC_n = \sum_{k=1}^n k \binom{n}{k}
\][/tex]
Substituting our earlier result:
[tex]\[
\sum_{k=1}^n k \binom{n}{k} = n \cdot 2^{n-1}
\][/tex]
Now let's consider the term [tex]\(\frac{1}{2}(n \cdot 2^n)\)[/tex]:
[tex]\[
\frac{1}{2}(n \cdot 2^n) = n \cdot 2^{n-1}
\][/tex]
Therefore, our original expression becomes:
[tex]\[
\sum_{k=1}^n k \binom{n}{k} - \frac{1}{2}(n \cdot 2^n) = n \cdot 2^{n-1} - n \cdot 2^{n-1} = 0
\][/tex]
Thus, we have shown the required result:
[tex]\[
C_1 + 2C_2 + 3C_3 + \cdots + nC_n - \frac{1}{2}(n \cdot 2^n) = 0
\][/tex]
