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Difference between revisions of "1977 AHSME Problems/Problem 10"

(Created page with "== Problem 10 == If <math>(3x-1)^7 = a_7x^7 + a_6x^6 + \cdots + a_0</math>, then <math>a_7 + a_6 + \cdots + a_0</math> equals <math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qqu...")
 
(No difference)

Latest revision as of 11:34, 21 November 2016

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 $(3(1)-1)^7) = \boxed{\text{(D)}\ 128}$.

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