Can anyone help me find the terms indicated in the expansions of this expression?

Question

(1-3p^2)^7 , p^10 term

i know you cna do this on calc, but is there a way to do it on paper?

Answer 1

With some work sure.

(1 - 3p^2)(1 - 3p^2)(1 - 3p^2)(1 - 3p^2)(1 - 3p^2)(1 - 3p^2)(1 - 3p^2)

And just work through all multiplications that give p^10

For example: 1 from the first and second terms and 3p^2 from the rest;

(1)(1)(3p^2)(3p^2)(3p^2)(3p^2)(3p^2) = 243p^10

And then how many combinations are there that give this:

Let me pair up the different terms that have the ones:

1/2, 1/3, 1/4, 1/5, 1/6, 1/7

2/3, 2/4, 2/5, 2/6, 2/7

3/4, 3/5, 3/6, 3/7

4/5, 4/6, 4/7

5/6, 5/7

6/7

So 21 of these.

21(243p^10) = 5103p^10

I think that does it. But I don't have a calculator to check it.

Answer 2

there is...

first  A = 1 B= -3p^2

use pascal's triangle to determine the coefficients for each term of the equation... anything raise to 7 would have coefficients equal to 1,7,21,35,35,21,7,1 these coefficients are for each of the terms from 1st term to last term

then...

[1]A^7 B^0 + [7] A^6 B^1  + [21] A^5 B^2 + [35] A^4 B^3 + [35] A^3 B^4 + [21] A^2 B^5 + [7] A^1 B^6 + [1] A^0 B^7

then substitute A with 1 and B with -3p^2

then find the term with p^10 :) [i believe that the term with p^ 10 is the 6th term because [21] A^2 B^5 = (21) (1)^2 (-3p^2)^5 = 21 (-243)(p^10)

(note the coefficients of each term are in brackets ->[ ])

note that the exponents of A are descending from 7->0 and the exponents of B are ascending from 0 -> 7

i hope that you get it because i know that at first glance the equation above might be O_O because of all the signs XP

determine A and B by making the expression = to A+B

so... (1-3p^2) = (1 + [-3p^2]) therefore A = 1 B = -3p^2