Expand it

The sum of the coefficients is simply the value when $x=y=1$ . We plug that in and see that the sum of the coefficients equals $(1+1)^9=2^9=\boxed{512}$ and we are done.

From the Binomial Theorem, we get

$(x+y)^n = \displaystyle \sum_{i=0}^n \binom {n}{i} x^{n-i} y^i$

Hence, the sum of the coefficients in $(x+y)^9$ is simply the sum of the binomial coefficients as follows: $\displaystyle \sum_{i=0}^9 {9 \choose i} = {9 \choose 0} + {9 \choose 1} + \ldots +{9 \choose 9} = 2^9 = \boxed{512}$ .

Generalisation

$\displaystyle \sum_{i=0}^n {n \choose i} = 2^n$

• Proof 1

Using the Binomial Theorem, we get $(1+1)^n = 2^n = \displaystyle \sum_{i=0}^n \binom {n}{i} 1^{n-i} 1^i = \displaystyle \sum_{i=0}^n \binom {n}{i}$

• Proof 2

The sum of the binomial coefficient is simply the number of subsets of the set { ${1,2,3, \ldots , n}$ } $= 2^n$ .

The solution here is to substitute $1$ to each of the variables then simplify. We have

$\large (x+y)^9$ $\implies$ $\large (1+1)^9$ $\implies$ $\large 2^9$ $\implies$ $\large 512$

$(x+y)^9$

Applying binomial theorem,

$\sum^9_i = 0$ ${9 \choose 1}$ $x^{(9-i)} y^i$

$=x^9 + 9x^8y + 36x^7y^2 + 84x^6y^3 + 126x^5y^4 +126x^4 y^5 + 84x^3 y^6 + 36x^2y^7+9xy^8+y^9$

Therefore the sum of coefficients,

$1 + 9 + 36 + 84 + 126 + 126 + 84 + 36 + 9 + 1 = \boxed{512}$

the pattern of is so easy. It's using pascal triangle.

and the sum for the first line of the pascal triangle is 1 $2^{0}$

second line is 2 $2^{1}$

third line is 4 $2^{2}$

So, if we want to find the sum of $(x + y)^{9}$ we use $2^{9}$ in the tenth line which equal to

$\boxed{512}$

2^9=512