Find the coefficient of x^5 in the expansion: I. (2+x)^7 II. (1+2x)^9?

Question

Hi. This is another propblem I'm having with the binomial theorem. According to my notes, I should just need to use my calculator to work out 7C5 for the first one...but its wrong...
the question:

Find the coefficient of x^5 in the expansion:
I. (2+x)^7
II. (1+2x)^9

Answer 1

well simply consider the formula we have:
(a+b)^n=sum((a^k)*(b^(n-k)*nCk) for k=0...n (1)
now in ur first question let a=2, b=x and n=7
by putting in (1) we can find the coeffcient of x^5
we have to let k=2 so by (1) :
(2^2)*(x^(7-2))*7C2 =4*168*x^5= 672x^5
so the answer is 672.of course i have to remark that 7C2=7C5 but you got the wrong answer because u didn't consider 2 in (2+x)^7 as a constant coeff...
in the second one by assuming k=4 by (1) we have: (a=1,b=2x,n=9 )so:
(1^4)*(2x)^5*9C4=32*126*x^5 so the answer is
32*162

Answer 2

1] One method is by expanding the brackets (there prob is an easier way but my algebra skills are abit rusty)
(2+x)^7
= (2+x) * (2+x) * (2+x) * (2+x) * (2+x) * (2+x) * (2+x)
= (4 + 4x + x^2) * (2+x) * (2+x) * (2+x) * (2+x) * (2+x)
= (8 + 12x + 6x^2 + x^3) * (2+x) * (2+x) * (2+x) * (2+x)
= (16 + 32x + 24x^2 + 8x^3 + x^4) * (2+x) * (2+x) * (2+x)
= (32 + 80x + 80x^2 + 40x^3 + 9x^4 + x^5) * (2+x) * (2+x)
= (64 + 192x + 240x^2 + 160x^3 + 58x^4 + 11x^5 + x^6) * (2+x)
= (128 + 448x + 672x^2 + 560x^3 + 276x^4 + 80x^5 + 13x^6 + x^7)

hence the coefficient of x^5 should be 80 (assuming i havent made a mistake along the way, just woke up so brains still a bit tired) ^_-

Answer 3

You forgot the rest of the expansion. 7C5 would be correct if it were (x + a)^7, but for (x + 2)^7, 2^2 becomes part of the coefficient of x^5:
(7C5)2^2 = 7*6*4/2 = 84

II.
Now you have
(9C5)2^5 = 9*8*7*6*32/(1*2*3*4)
or, if you prefer,
9!2^5/(5!4!) = 4,032

Answer 4

P(x) = (x + a million)(x + 2)(x + 3) . . . (x + n) employing kin between roots ( 0) and coefficients ( Viete or Vieta ) x^n has coefficient a million x^(n-a million) has coefficient a million+2 + 3+ ....+ n = n(n+a million)/2 x^(n-2) has coefficient a million*2+ a million*3+ .. a million*n + 2*3+ .. 2*n +..... (n-a million)*n it extremely is sum of any double products x_i*x_j the place i < j for i between a million,n-a million and j between 2 and n x^(n-r) is sum of goods x_i_1 * x_i_2* .... *x_i_r the place i_1< i_2