Question
The binomial expansion of (5-x/2)^6 is 15625 + px+qx^2+... Find the value of "p" and the value of "q"
Answers
Reiny
(5-x/2)^6
I assume you know how to expand using the binomial theorem.
(5-x/2)^6
= 5^6 + 5^5(-x/2) + 5^4(-x/2)^2 + ...
= 15625 - (3125/2)x + (625/4)x^2 + ...
so matching up terms:
-3125x/2 = px ----> p = -3125/2
(625/4)x^2 = qx^2 ----> q = 625/4
I assume you know how to expand using the binomial theorem.
(5-x/2)^6
= 5^6 + 5^5(-x/2) + 5^4(-x/2)^2 + ...
= 15625 - (3125/2)x + (625/4)x^2 + ...
so matching up terms:
-3125x/2 = px ----> p = -3125/2
(625/4)x^2 = qx^2 ----> q = 625/4