To find the coefficient of the x^5 term of the binomial expansion, we should recall the Binomial Theorem. It states that for every positive integer n, we can expand the expression (a+b)^n by using the numbers in the n^(th) row of Pascal's Triangle.
cc
(a+b)^n= & _nC_0a^nb^0+ _nC_1a^(n-1)b^1
& +
& ...
& +
& _nC_(n-1)a^1b^(n-1)+ _nC_na^0b^n
In the above formula, _nC_0, _nC_1, ..., _nC_n are the numbers in the n^(th) row of Pascal's Triangle.
Row 1.3cm Pascal's Triangle 1.2cm
cccccccccccc
0 & & & & & & 1 & & & & &
1 & & & & & 1 & & 1 & & & &
2 & & & & 1 & & 2 & & 1 & & &
3 & & & 1 & & 3 & & 3 & & 1 & &
4 & & 1 & & 4 & & 6 & & 4 & & 1 &
5 & 1 & & 5 & & 10 & & 10 & & 5 & & 1Note that each number greater than 1 found in the triangle is the sum of the two numbers diagonally above it. Now consider the given binomial.
(x-2)^(10) ⇔ ( x+( -2))^(10)
We can substitute the first term for a and the second term for b using the Binomial Theorem equation.
Notice that each term in the expansion has the form _(10)C_r x^(10- r)( -2)^r. From this we can tell that the term containing x^5 occurs when r= 5. Let's start by evaluating _(10)C_5. To do so, recall the formula for the number of combinations of n objects taken r at a time, where r≤ n.
_n C_r=n!/(n-r)! r!
Keeping this in mind, let's evaluate _(10)C_5 by substituting n = 10 and r = 5 into the formula.
