# 1977 AHSME Problems/Problem 10

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem 10

If $(3x-1)^7 = a_7x^7 + a_6x^6 + \cdots + a_0$, then $a_7 + a_6 + \cdots + a_0$ equals

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 64 \qquad \text{(D)}\ -64 \qquad \text{(E)}\ 128$

## Solution

Solution by e_power_pi_times_i

Notice that if $x=1$, then $a_7x^7 + a_6x^6 + \cdots + a_0 = a_7 + a_6 + \cdots + a_0$. Therefore the answer is $(3(1)-1)^7) = \boxed{\text{(D)}\ 128}$.