The sum of the series $\displaystyle\sum_{r=0}^{10} {}^{20}C_r$ is $2^{19} + \dfrac{{}^{20}C_{10}}{2}$.

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> [!TLDR]
> The sum of binomial coefficients $\sum_{r=0}^{10} {}^{20}C_r$ is found by leveraging the symmetry property ${^nC_r} = {^nC_{n-r}}$ and the total sum $\sum_{r=0}^{n} {^nC_r} = 2^n$. The calculation confirms the given statement is true.

The problem asks us to verify the sum of a specific part of a binomial expansion. To approach this, we need to recall the fundamental properties of binomial coefficients, particularly their sum and symmetry.

The Binomial Theorem provides the expansion of $(a+b)^n$ as $\sum_{r=0}^n {^nC_r a^{n-r} b^r}$. A special case arises when $a=1$ and $b=1$:
$(1+1)^n = \sum_{r=0}^n {^nC_r (1)^{n-r} (1)^r}$
$2^n = \sum_{r=0}^n {^nC_r}$

This means the sum of all binomial coefficients for a given $n$ is $2^n$. For our problem, $n=20$, so the sum of all coefficients from $r=0$ to $r=20$ is $2^{20}$.

Another crucial property is the symmetry of binomial coefficients:
${^nC_r} = {^nC_{n-r}}$
This property tells us that coefficients equidistant from the beginning and end of the expansion are equal. For $n=20$, this means ${}^{20}C_0 = {}^{20}C_{20}$, ${}^{20}C_1 = {}^{20}C_{19}$, and so on. The middle term, when $n$ is even, is ${^nC_{n/2}}$, which in our case is ${}^{20}C_{10}$.

We are asked to find the sum $\sum_{r=0}^{10} {}^{20}C_r = {}^{20}C_0 + {}^{20}C_1 + \dots + {}^{20}C_{10}$. Let's use these properties to evaluate this sum.

1.  **Identify the total sum:**
    For $n=20$, the sum of all binomial coefficients is $2^{20}$.
    $$ \sum_{r=0}^{20} {}^{20}C_r = {}^{20}C_0 + {}^{20}C_1 + \dots + {}^{20}C_{10} + {}^{20}C_{11} + \dots + {}^{20}C_{20} = 2^{20} $$

2.  **Break down the total sum using symmetry:**
    Let $S = \sum_{r=0}^{10} {}^{20}C_r = {}^{20}C_0 + {}^{20}C_1 + \dots + {}^{20}C_{10}$.
    We can rewrite the full sum as:
    $$ \sum_{r=0}^{20} {}^{20}C_r = \left( {}^{20}C_0 + \dots + {}^{20}C_9 \right) + {}^{20}C_{10} + \left( {}^{20}C_{11} + \dots + {}^{20}C_{20} \right) $$
    Let $A = {}^{20}C_0 + {}^{20}C_1 + \dots +

Power	Expansion
$(x + y)^0$	$1$
$(x + y)^1$	$x + y$
$(x + y)^2$	$x^2 + 2xy + y^2$
$(x + y)^3$	$x^3 + 3x^2y + 3xy^2 + y^3$
$(x + y)^4$	$x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$

$k$	$\binom{4}{k}$	$x^{4-k}$	$y^k$	Term
0	$\binom{4}{0}=1$	$x^4$	$y^0$	$1 \cdot x^4$
1	$\binom{4}{1}=4$	$x^3$	$y^1$	$4x^3y$
2	$\binom{4}{2}=6$	$x^2$	$y^2$	$6x^2y^2$
3	$\binom{4}{3}=4$	$x^1$	$y^3$	$4xy^3$
4	$\binom{4}{4}=1$	$x^0$	$y^4$	$1 \cdot y^4$

Q.The sum of the series $\displaystyle\sum_{r=0}^{10} {}^{20}C_r$ is $2^{19} + \dfrac{{}^{20}C_{10}}{2}$ .

Concept understanding — Binomial Theorem Expansion

The Problem That Started It All

The Pattern You Already Know

The Precise Statement

Where Do Those Coefficients Come From?

The General Term

What If It's Not Just $x$ and $y$ ? …

Q.The sum of the series ∑r=01020Cr\displaystyle\sum_{r=0}^{10} {}^{20}C_rr=0∑10​20Cr​ is 219+20C1022^{19} + \dfrac{{}^{20}C_{10}}{2}219+220C10​​.

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Q.The sum of the series $\displaystyle\sum_{r=0}^{10} {}^{20}C_r$ is $2^{19} + \dfrac{{}^{20}C_{10}}{2}$ .

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