Given that$(3x-1)^7=a_7x^7+a_6x^6+a_5x^5+...+a_1x+a_0$, find $a_7+a_6+a_5+a_4+...+a_1+a_0$ Given that$$(3x-1)^7=a_7x^7+a_6x^6+a_5x^5+...+a_1x+a_0$$find $$a_7+a_6+a_5+a_4+...+a_1+a_0$$
Is there anyway to do the question without using binomial theorem and expanding the expression on the LHS??

Consider what happens to $\sum_{i=0}^{7} a_i x^i$ when using $x = 1$. You get $\sum_{i=0}^{7} a_i$. So, calculate $(3x - 1)^7$ when $x = 1$ which is $2^7 = 128$.
