Question

(32C0)^2-(32C1)^2+(32C2)^2-..........(32C32)^2=

Answer 1

Answer:

263-1/2(64C32)

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Answer 2

SOLUTION

TO DETERMINE

$\displaystyle \sf{ \bigg({{}^{32}C_0 }\bigg)^{2} - \bigg({{}^{32}C_1 }} \bigg)^{2} + \bigg({{}^{32}C_2 } \bigg)^{2} + ... + \bigg({{}^{32}C_{32} } \bigg)^{2}$

EVALUATION

$\displaystyle \sf{ \bigg({{}^{32}C_0 }\bigg)^{2} - \bigg({{}^{32}C_1 }} \bigg)^{2} + \bigg({{}^{32}C_2 } \bigg)^{2} + ... + \bigg({{}^{32}C_{32} } \bigg)^{2}$

$= \displaystyle \sf{ \sum\limits_{r=0}^{n}{( - 1)}^{r} \bigg({{}^{32}C_{r} } \bigg)^{2}}$

$= \displaystyle \sf{ \sum\limits_{r=0}^{n} {( - 1)}^{r} \bigg({{}^{32}C_{r} } \bigg)}\bigg({{}^{32}C_{r} } \bigg)$

$= \displaystyle \sf{ \sum\limits_{r=0}^{n} {( - 1)}^{r} \bigg({{}^{32}C_{r} } \bigg)}\bigg({{}^{32}C_{32 - r} } \bigg) \: \because \{ {}^{n}C_{r} = {}^{n}C_{n - r} \}$

Now the suffixes are r and 32 - r

So the sum of the suffixes = 32 - r + r = 32

Hence the given expression

$= \sf{Coefficient \: of \: {x}^{32} \: in \: {(1 - x)}^{32} {(1 + x)}^{32} }$

$= \sf{Coefficient \: of \: {x}^{32} \: in \: {(1 - {x}^{2} )}^{32} }$

$= \sf{Coefficient \: of \: {( {x}^{2} )}^{16} \: in \: {(1 - {x}^{2} )}^{32} }$

$= \displaystyle \sf{ {{}^{32}C_{16}}}$

FINAL ANSWER

$\displaystyle \sf{ \bigg({{}^{32}C_0 }\bigg)^{2} - \bigg({{}^{32}C_1 }} \bigg)^{2} + \bigg({{}^{32}C_2 } \bigg)^{2} + ... + \bigg({{}^{32}C_{32} } \bigg)^{2} = \displaystyle \sf{ {{}^{32}C_{16}}}$

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