How do you find the coefficient of #a^2# in the expansion of #(2a+1)^5#?

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Answer 1 · 2016-07-04T19:35:22

Coefficient of #a^2# is 40.

Binomial theorem is

#(x+y)^n = sum_(k=0)^n ((n),(k))x^(n-k)y^k#

We want the a^2 term so if x = 2a then we need #n-k = 2#, since n = 5 this implies that k = 3. Hence the term we are interested in is

#((5),(3))(2a)^(5-3)(1)^3#

#(5!)/(3!2!)*(2a)^2*1^3 = (5*4*(cancel(3*2*1)))/((cancel(3*2*1))(2*1))*4a^2 = 10*4a^2 = 40a^2#

Of course, you could also expand the whole thing out using Pascal's triangle or other methods but this is the easiest way in my opinion.

Answer 2 · 2016-07-04T20:29:39

#40#

#p_n(a)=(2a+1)^n# is a polynomial in #a#

If #b_k# is the coefficient of #a^k# it can be obtained by calculating

#b_k = 1/(k!)(d^k)/(da^k)p_n(a)_{a = 0}#

in the present case

#b_2=1/(2!)(5 xx 4)2^2(2xx0+1)^3=40#