Differentiation form of Binomial Coefficients - Practice Questions & MCQ

1. To get the value of $C_1+2 C_2+3 C_3+\ldots \ldots \ldots+n C_n$

We know, $(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots \ldots+C_r x^r+\ldots \ldots+C_n x^n$
Differentiate both sides concerning the x we get,

$
n(1+x)^{n1}=0+C_1+2 C_2 x+3 C_3 x^2+\ldots .+n C_n x^{n1}
$
Put $x=1$ we get,

$
n \cdot 2^{n1}=C_1+2 C_2+3 C_3+\ldots \ldots+n C_n
$
Hence, $C_1+2 C_2+3 C_3+\ldots \ldots+n C_n=n .2^{n1}$
Note:
This result can also be written as $\sum_{r=0}^n r .{ }^n C_r=\sum_{r=1}^n r .{ }^n C_r=n \cdot 2^{n1}$
Students are advised to remember this result

Example1
The value of $C_0+2 C_1+3 C_2+\ldots \ldots+(n+1) C_n$ is

Method 1: Differentiation

$
x(1+x)^n=C_0 x+C_1 x^2+C_2 x^3+\ldots \ldots+C_n x^{n+1}
$
Differentiate both sides concerning the x we get,

$
x \cdot n(1+x)^{n1}+(1+x)^n \cdot 1=C_0+2 C_1 x+3 C_2 x^2+\ldots \ldots+(n+1) C_n x^n
$
Put $x=1$ we get,

$
\begin{aligned}
& n(2)^{n1}+2^n=C_o+2 C_1+3 C_2+\ldots \ldots+(n+1) C_n \\
& C_0+2 C_1+3 C_2+\ldots \ldots+(n+1) C_n=(n+2) 2^{n1}
\end{aligned}
$
Method 2: Sigma method

$
\begin{aligned}
& C_0+2 C_1+3 C_2+\ldots \ldots+(n+1) C_n \\
& =\sum_{r=0}^n(r+1) \cdot{ }^n C_r \\
& =\sum_{r=0}^n r .{ }^n C_r+\sum_{r=0}^n{ }^n C_r \\
& =n \cdot 2^{n1}+2^n \quad \text { (Using result derived in previous section) } \\
& =2^{n1}(n+2)
\end{aligned}
$

Example 2

$
C_0+3 C_1+5 C_2+\ldots+(2 n+1) C_n=?
$
Method 1: Differentiation
The given series is

$
(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots+C_n x^n
$
Now, replacing $x$ by $x^2$, we get

$
\left(1+x^2\right)^n=C_0+C_1 x^2+C_2 x^4+\ldots+C_n x^{2 n}
$
On multiplying both sides by $x^1$, we get

$
x\left(1+x^2\right)^n=C_0 x+C_1 x^3+C_2 x^5+\ldots+C_n x^{2 n+1}
$
On differentiating both sides w.r.t x

$
x \cdot n\left(1+x^2\right)^{n1} \cdot 2 x+\left(1+x^2\right)^n \cdot 1=C_0+3 C_1 x^2+5 C_2 x^4+\ldots+(2 n+1) C_n x^{2 n}
$
Putting $x=1$, we get

$
n \cdot 2^{n1} \cdot 2+2^n=C_0+3 C_1+5 C_2+\ldots+(2 n+1) C_n
$
Hence, $C_0+3 C_1+5 C_2+\ldots+(2 n+1) C_n=(n+1) 2^n$.

Method 2: Sigma method

$
\begin{aligned}
& C_0+3 C_1+5 C_2+\ldots+(2 n+1) C_n \\
& =\sum_{r=0}^n(2 r+1) \cdot{ }^n C_r \\
& =2 \sum_{r=0}^n r \cdot{ }^n C_r+\sum_{r=0}^n{ }^n C_r \\
& =2 \cdot n \cdot 2^{n1}+2^n \\
& =(n+1) \cdot 2^n
\end{aligned}
$
Important Results for Sigma Method
1. $\sum_{r=0}^n{ }^n C_r=2^n$
2. $\sum_{r=0}^n r .{ }^n C_r=\sum_{r=1}^n r .{ }^n C_r=n .2^{n1}$
$\sum_{\text {3. }}^{n=0} r^2 \cdot{ }^n C_r=\sum_{r=1}^n r^2 \cdot{ }^n C_r=n(n+1) 2^{n2}$