Please help with this maths question? Binomials

10) Using the first three terms of the expansion of (1 – 2x)¹² and the substitution x = 0.01 gives an approximate value for 0.98¹² of...

(a) 0.7847
(b) 0.7864
(c) 0.7732
(d) 0.7336
(e) I don’t know

Can someone explain this to me, please? I get all the other questions but not this one.

Thanks :)

The general expression for the term:
................ n
(a + b)^n = ∑ nCk a^k b^(n-k); where nCk = n! / ( k! (n-k)! )
................ k=0

Thus the first three terms (k = 0, 1, 2), for (1-2x)^12 would be:

= 12! / ( 0! (12-0)! ) 1^0 (-2x)^(12-0)
+ 12! / ( 1! (12-1)! ) 1^1 (-2x)^(12-1)
+ 12! / ( 2! (12-2)! ) 1^2 (-2x)^(12-2)

= (-2x)^12 + 12 (-2x)^(11) + 66(-2x)^(10)
= (-2x)^10 [ (-2x)^2 + 12 (-2x) + 66 ]
= (-2x)^10 [ 4x^2 - 24x + 66 ]

@ x = .01
= (-2*.01)^10 [ 4(.01)^2 - 24(.01) + 66 ]
= 6.73386496 x 10(-16) ← this is for the first 3 terms

...

For the LAST three terms (k = n, n-1, n-2), for (1-2x)^12 would be:

= 1 (-2x)^0 + 12 (-2x)^1 + 66 (-2x)^2

@ x = .01
1 (-2*.01)^0 + 12 (-2*.01)^1 + 66 (-2*.01)^2
= .7864 ← this is for the last 3 terms

Why would you want to do that when there are so many nice calculators to perform the computation?

The expansion you requested contains 13 terms. That is a lot of work to compute the value of 13 terms and then add them.

To find those 13 terms use the web source below. I suggest you form Pascal's Triangle (it is not difficult) to the 12 level to get the coefficients of the expansion. Good Luck.

Answer is a). Yes, I used a calculator.

(a) 0.7847