- #1

RChristenk

- 57

- 7

- Homework Statement
- Find the sum of the coefficients of ##(x+y)^{16}##

- Relevant Equations
- Binomial Theorem

##(x+y)^{16}=[x(1+\dfrac{y}{x})]^{16}=x^{16}(1+\dfrac{y}{x})^{16}##

## x^{16}(1+\dfrac{y}{x})^{16}=x^{16}[1+^{16}C_1(\dfrac{y}{x})+^{16}C_2(\dfrac{y}{x})^2...+^{16}C_{16}(\dfrac{y}{x})^{16}]##

Now let ##x=1,y=1##:

##1^{16}(1+1)^{16}=1^{16}(1+^{16}C_1+^{16}C_2...+^{16}C_{16})##

##2^{16}-1=^{16}C_1+^{16}C_2...+^{16}C_{16}##

Sum of coefficients = ##2^{16}-1## = ##65535##

But the answer is ##65536##. Why?

## x^{16}(1+\dfrac{y}{x})^{16}=x^{16}[1+^{16}C_1(\dfrac{y}{x})+^{16}C_2(\dfrac{y}{x})^2...+^{16}C_{16}(\dfrac{y}{x})^{16}]##

Now let ##x=1,y=1##:

##1^{16}(1+1)^{16}=1^{16}(1+^{16}C_1+^{16}C_2...+^{16}C_{16})##

##2^{16}-1=^{16}C_1+^{16}C_2...+^{16}C_{16}##

Sum of coefficients = ##2^{16}-1## = ##65535##

But the answer is ##65536##. Why?