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

  • Thread starter RChristenk
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  • #1
RChristenk
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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?
 
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  • #2
RChristenk said:
Sum of coefficients = ##2^{16}-1## = ##65535##
The first '=' in that line is not correct. It does not follow from anything written above it.
You don't need most of the working in the OP. The sum of the coefficients will simply be the value of the expression when ##x=y=1##, since all items ##x^k y^{16-k}## will be 1. Hence the sum of the coefficients will just be ##(1+1)^{16}##.
 
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  • #3
andrewkirk said:
The first '=' in that line is not correct. It does not follow from anything written above it.
You don't need most of the working in the OP. The sum of the coefficients will simply be the value of the expression when ##x=y=1##, since all items ##x^k y^{16-k}## will be 1. Hence the sum of the coefficients will just be ##(1+1)^{16}##.
Could you tell me where I went wrong specifically? Because to me what I wrote down looks correct (although obviously it isn't). I know I could just set everything to 1 and plug it in, but then I don't really understand what's happening and it becomes a rote memory item to me.
 
  • #4
RChristenk said:
Could you tell me where I went wrong specifically? Because to me what I wrote down looks correct (although obviously it isn't). I know I could just set everything to 1 and plug it in, but then I don't really understand what's happening and it becomes a rote memory item to me.
You moved the first coefficient to the left hand side and gave the answer as ##N -1## rather than ##N##. Only you can explain why you did this!
 
  • #5
To rephrase @PeroK #4 :

RChristenk said:
Could you tell me where I went wrong specifically?

You overlooked that this number one is also a coefficient (##^{16}C_0## )

(easy check: same exercise with powers 0, 1, 2, ... instead of 16 :smile:)

##\ ##
 
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  • #6
PeroK said:
You moved the first coefficient to the left hand side and gave the answer as ##N -1## rather than ##N##. Only you can explain why you did this!
Uh..I thought ##C## represented the "C"oefficients, so naturally the digit one is moved to the other side. Now I see that is incorrect.
 
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