Summing it up !!!

$(1-x)^{10}$ coefficients are equal to $(1-1)^{10}$ coefficients.

Now, developed the last expression, we get it:

$(1-1)^{10}=\binom{10}{0}-\binom{10}{1}+\binom{10}{2}...+\binom{10}{10}$

$(0)^{10}=\binom{10}{0}-\binom{10}{1}+\binom{10}{2}...+\binom{10}{10}$

$0 =\binom{10}{0}-\binom{10}{1}+\binom{10}{2}...+\binom{10}{10}$

$\binom{10}{1}+\binom{10}{3}+...+\binom{10}{9}=\binom{10}{0}+\binom{10}{2}...+\binom{10}{10}$

Of that we conclude that negative sum coeffficients are equal to negative positive coeffficients. So sum of al positive coefficients are equal to the sum of all coefficientes divided two.

The sum of al coefficientes is $2^n$ , demostration is similiar but we use $(1+1)^{10}$ .

Finally sum is $\frac{2^{10}}{2}=2^9=\boxed{512}$

For some reason I do not like binomial theorem and I used pascal's triangle because it was easier in this case. Since you have very easy numbers (you're not multiplying by anything more than 1 or negative 1) you take the sum of every other number in the 10th row of pascal's triangle because they are the coefficients for each term and the first term is positive. That's 1 + 45 + 210 + 210 + 45 + 1, which gives us our answer 512.

First I will be creating a pattern, and then will solve the problem. In expansion of ${ (1-x) }^{ 10 }$ , we get coeffecients such as ${ { 10 }_{ C } }_{ 1 },\quad { { 10 }_{ C } }_{ 2 },\quad and\quad { { 10 }_{ C } }_{ 10 }.$ Clearly, by writing the first three values, we ca see that the alternate terms, starting from ${ { 10 }_{ C } }_{ 1 }$ are even. So, we need to find ${ { 10 }_{ C } }_{ 2 }+{ { 10 }_{ C } }_{ 3 }+{ { 10 }_{ C } }_{ 5 }+{ { 10 }_{ C } }_{ 7 }+{ { 10 }_{ C } }_{ 9 }$

Now, we will calculate this value: $Using\quad Binomial\quad Theorem,\\ { (1+x) }^{ 10 }\quad =\quad 1\quad +{ { \quad 10 }_{ C } }_{ 1 }x\quad +{ { \quad 10 }_{ C } }_{ 2 }{ x }^{ 2 }\quad +....+\quad { { 10 }_{ C } }_{ 10 }{ x }^{ 10 }\\ Now,\quad putting\quad x=1\\ { 2 }^{ 10 }\quad =\quad 1\quad +\quad { { 10 }_{ C } }_{ 1 }\quad +\quad { { 10 }_{ C } }_{ 2 }\quad +......+\quad { { 10 }_{ C } }_{ 10 }\quad -----1.\\ \\ Putting\quad x=-1,\\ 0\quad =\quad 1\quad -\quad { { 10 }_{ C } }_{ 1 }\quad +\quad { { 10 }_{ C } }_{ 2 }\quad -.........+\quad { { 10 }_{ C } }_{ 10 }\quad ------2.\\ \\ Subtracting\quad 2.\quad from\quad 1.\\ { 2 }^{ 10 }\quad =\quad 2({ { 10 }_{ C } }_{ 1 }+{ { 10 }_{ C } }_{ 3 }.....{ { 10 }_{ C } }_{ 9 })\\ { { 10 }_{ C } }_{ 1 }+{ { 10 }_{ C } }_{ 3 }.....{ { 10 }_{ C } }_{ 9 }\quad =\quad { 2 }^{ 9 }\\ { { 10 }_{ C } }_{ 1 }+{ { 10 }_{ C } }_{ 3 }.....{ { 10 }_{ C } }_{ 9 }\quad =\quad 512$

Although people below have given good solutions to this answer, it is worth noting that for the binomial expansion of a term such as $(x - 1)^n$ , where $n$ is a whole number, the following rule holds true:

The sum of all the coefficients of the terms of the sequence is $2^n$ and, the sum of the coefficients of the positive terms is equal to the sum of the coeffients of the negative terms which is equal to $2^{n-1}$

2^(n-1) where n is the power of the expression if it is asked to find the sum of all the terms, suppose in case of (x+1)^10, then the answer would be 2^n.