Sum of Binomial Coefficients

The Sum of Binomial Coefficient is an important concept of algebra that helps to expand the expressions. A Binomial is an expression with two terms. It is difficult to solve the powers manually therefore this expression makes it simpler to solve. This theorem is widely used in real-life applications in mathematics including calculus etc.

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Binomial Expression:

An algebraic expression consisting of only two terms is called a Binomial Expression

$
e g \cdot(a+b)^2,\left(\sqrt{x}+\frac{k}{x^2}\right)^5,(x+9 y)^{-2 / 3}
$

If we wanted to expand $(x+y)^{52}$, we might multiply $(x+y)$ by itself fifty-two times. This could take hours!
But if we examine some simple expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.

$
\begin{aligned}
& (x+y)^2=x^2+2 x y+y^2 \\
& (x+y)^3=x^3+3 x^2 y+3 x y^2+y^3 \\
& (x+y)^4=x^4+4 x^3 y+6 x^2 y^2+4 x y^3+y^4
\end{aligned}
$

On examining the exponents, we find that with each successive term, the exponent for x decreases by 1 and the exponent for y increases by $1$ . The sum of the two exponents is $n$ for each term.
Also the coefficients for $(x+y)^n$ are equal to $\binom{n}{0},\binom{n}{1},\binom{n}{2}, \ldots,\binom{n}{n}$ where, $\binom{n}{r}=C(n, r)={ }^n C_r=\frac{n!}{r!(n-r)!}$

These patterns lead us to the Binomial Theorem, which can be used to expand any binomial expression.

Binomial Theorem

If $n$ is any positive integer, then

$ (a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{n - 1} b + \binom{n}{2} a^{n - 2} b^2 + \dots + \binom{n}{n - 1} a b^{n - 1} + \binom{n}{n} b^n $

Binomial Coefficient

The combination $\binom{n}{r}$ or ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}$ occuring in the Binomial theorem is called a Binomial coefficient, where $\binom{n}{r}=C(n, r)={ }^n C_r=\frac{n!}{r!(n-r)!}$.

Some Standard Expansions on Sum of Binomial Coefficients:

1. Sum of Binomial Coefficients

$
\begin{aligned}
& C_0+\mathrm{C}_1+C_2+C_3+\ldots \ldots+C_n=2^n \\
& \text { or } \sum_{r=0}^n{ }^n C_r=2^n
\end{aligned}
$

2. Sum of Binomial coefficients with alternate sign

The sim of Binomial coefficeints with alternate sign are zero as the terms $T_k$ and $T_{n-k}$ are equal.

$
\begin{aligned}
& C_0-\mathrm{C}_1+C_2-C_3+\ldots \ldots+(-1)^n C_n=0 \\
& \text { or } \quad \sum_{r=0}^n(-1)^r{ }^n C_r=0
\end{aligned}
$

3. Sum of the Binomial coefficients of the odd terms / Sum of the Binomial coefficients of the even terms

The sum of the binomial coefficients of the odd terms = Sum of the binomial coefficients of the even terms

$
\text { I.e. } C_1+C_3+C_5 \ldots \ldots=C_0+C_2+C_4+\ldots \ldots=2^{n-1}
$
solve and analyze real-life complex problems.

Summary

Binomial theorem is used for the expansion of a binomial expression with a higher degree. Binomial coefficients are the coefficients of the terms in the Binomial expansion. Understanding the sum of two binomial coefficients gives an idea to solve more complex problems in calculus, statistics, data analysis etc.

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Solved Examples Based on Sum of Binomial Coefficients

Example 1: Let $S_n=1+q+q^2+\ldots \ldots \ldots+q^n$ and $T_n=1+\left(\frac{q+1}{2}\right)+\left(\frac{q+1}{2}\right)^2+\ldots \ldots \ldots \ldots+\left(\frac{q+1}{2}\right)^n$ where q is a real numer and $q \neq 1$ ${ }_{\text {If }}{ }^{101} C_1+{ }^{101} C_2 \cdot S_1+\ldots \ldots . .+{ }^{101} C_{101} \cdot S_{100}=\alpha T_{100}$ then $\alpha$ is equal to :
1) 200
2) $2^{100}$
3) 2021
4) $2^{99}$

Solution
Sum of Binomial Coefficients

$
C_0+C_1+C_2+C_3+----+C_n=2^n
$
$\begin{aligned}
&T_n=1+\left(\frac{q+1}{2}\right)+\left(\frac{q+1}{2}\right)^2+\cdots+\left(\frac{q+1}{2}\right)^n=\frac{\left(\frac{q+1}{2}\right)^{n+1}-1}{\left(\frac{q+1}{2}\right)-1}\\
&\text { Also, given, }\\
& { }^{101} \mathrm{C}_1+{ }^{101} \mathrm{C}_2 S_1+\ldots \ldots \ldots \ldots+{ }^{101} \mathrm{C}_{101} S_{100}=\alpha T_{100} \\
& \alpha T_{100}=\sum_{r=1}^{101}{ }^{101} C_r \cdot S_{r-1} \\
& \alpha T_{100}=\sum_{r=1}^{101}{ }^{101} C_r\left[\frac{q^r-1}{q-1}\right] \\
& \alpha T_{100}=\frac{1}{q-1}\left(\sum_{r=1}^{101}{ }^{101} C_r\left(q^r-1\right)\right) \\
& \alpha T_{100}=\frac{1}{q-1}\left((1+q)^{101}-1-\left(2^{101}-1\right)\right) \\
& \alpha T_{100}=\frac{1}{q-1}\left((1+q)^{101}-2^{101}\right)
\end{aligned}$

$\begin{aligned}
& \alpha\left(\frac{\left(\frac{q+1}{2}\right)^{101}-1}{\left(\frac{q+1}{2}\right)-1}\right)=\frac{(1+q)^{101}-2^{101}}{q-1} \\
& \frac{2}{2^{101}} \alpha\left(\frac{(q+1)^{101}-2^{101}}{(q+1)-2}\right)=\frac{(1+q)^{101}-2^{101}}{q-1} \\
& \alpha\left(\frac{1}{2^{100}}\right)=1 \\
& \alpha=2^{100}
\end{aligned}$

Example 2: If the number of terms in the expansion of $\left(1-\frac{2}{x}+\frac{4}{y}\right)^n, x, y \neq 0$ is 28 , then the sum of the coefficients of all the terms in this expansion is
1) $64$
2) $2187$
3) $243$
4) $729$

Solution:

$
\left(1-\frac{2}{x}+\frac{4}{y}\right)^n
$
Using multinomial theorem, number of terms ${ }^{n+2} C_2=28$

$
\begin{aligned}
& \frac{(n+2)!}{2!((n+2)-2)!}=28 \\
& \frac{(n+2)(n+1) n!}{2!n!}=28 \\
& (n+2)(n+1)=56
\end{aligned}
$

Thus $n=6$
So, in $\left(1-\frac{2}{x}+\frac{4}{y}\right)^6$
Put $x=1, y=1$
We get sum of coefficients $=3^6=729$

Example 3: If the fractional part of the number $\frac{2^{403}}{15}$ is $\frac{k}{15}$, then k is equal to:
1) $6$
2) $ 8$
3) $4$
4) $14$

Solution
Sum of Binomial Coefficients

$
C_0+C_1+C_2+C_3+----+C_n=2^n
$
Now,
$2^{403}$ can be written as

$
2^{403}=2^3 \cdot 2^{400}=8\left(2^4\right)^{100}=8(15+1)^{100}
$

So,

$
\begin{aligned}
\Rightarrow \frac{8}{15}(15+1)^{100} & =\frac{8}{15}(15 \lambda+1) \\
& =8 \lambda+\frac{8}{15}
\end{aligned}
$

$\because 8 \lambda$ is integer and $\frac{8}{15}$ is fractional part
So, $k=8$

Example 4: What is the sum of the coefficient of the term $(\sqrt{2}-\sqrt[3]{3}+\sqrt[6]{5})^{10}$ ?
1) $3^{10}$
2) $2^{10}$
3) 1
4) None of the above

Solution
Series Involving Binomial Coefficients -
In the expansion of $(x+y+z)^n$ if we put $x=y=z=1$, then we get the sum of coefficients. In this case, $(1+1+1)^n=\underline{3^n}$.
Sum of the coefficient $\left(x_1+x_2+x_3\right)^n=(1+1+1)^n=3^n$
To determine the coefficient
Put $x_1=1, x_2=-1, x_3=1$ and $n=10$
$(1-1+1)^n=1^n=1$
option C is correct.

Example 5: If $C_r={ }^{25} \mathrm{C}_r$ and $C_0+5 \cdot C_1+9 \cdot C_2+\cdots \cdots+(101) \cdot C_{25}=2^{25} \cdot k$, then k is equal to $\qquad$
1) 15
2) 30
3) 51
4) 27

Solution
The general term for the given series is: $(4 \mathrm{r}+1) .{ }^{25} \mathrm{C}_{\mathrm{r}}$
Applying summation
是
$=4 \sum_{r=1}^{25} r \times{ }^{25} C_r+2^{25}$
$=4 \times 25 \cdot 2^{24}+2^{25}=50 \cdot 2^{25}+2^{25}=(50+1) 2^{25}=51 \cdot 2^{25}$
$k=51$
Hence, the answer is the option 3.