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Binomial Theorem to find 1.16
By Alex Shandro on Saturday, October 06, 2001 - 02:43 pm:

I am stuck on some portfolio work for the International Baccalaureate. In particular I am stuck on two questions.

  1. By letting x=0.1 and using the expansion for (1+x)6, find the value of 1.16 without using a calculator, and confirm the result.
  2. Using the expansion of (1-x)n, and the expansion of (a+x)n, where a is a positive integer, evaluate, without using a calculator: (i) 0.996 (ii) 3.025. Also find the term in x3, in the expansion of (3+2x)5.
By Dan Goodman on Saturday, October 06, 2001 - 03:29 pm:

Do you know the binomial expansion (x+y)n? It is

(x+y)n = xn + nxn-1y + (n2)xn-2y2 + (n3)xn-3y3 + ... + yn

where (nr)=n!/(r!(n-r)!) and r!=r(r-1)(r-2)...(3)(2)(1).

So, (1+x)6=1+6x+15x2+20x3+15x4+6x 5+x6.

Now, set x=0.1, so you can find (1+0.1)6.
So x2=0.01, x3=0.001, etc.
So 6x=0.6, 15x2=0.15, etc.
Add these together to find 1.16.

Hopefully that should give you the idea for the second one as well, let us know if you still can't do it.