Binomial Expansions
This set continues on with all the work already studied on Pascal's Triangle, Combinations, and Binomial Expansions and Coefficients.
The questions here are more difficult, but the difficulty only adds an extra layer of thought. Take it slowly and go step by step. The concepts are the same but the way you might apply them now might be more complex.
The concepts we have covered include:
- What we get when we combine them is the knowledge that for an expansion of $(a+b)^n$(a+b)n the coefficients are dictated by the combinations of $\nCr{n}{0},\nCr{n}{1},\nCr{n}{2}$nC0,nC1,nC2 $...$... $\nCr{n}{n}$nCn
- So this results in the expansion looking like this
$(a+b)^n$(a+b)n
$=^nC_0a^n+^nC_1a^{n-1}b^1+^nC_2a^{n-2}b^2+....^nC_ra^{n-r}b^r+....^nC_{n-1}a^1b^{n-1}+^nC_nb^n$=nC0an+nC1an−1b1+nC2an−2b2+....nCran−rbr+....nCn−1a1bn−1+nCnbn
- Thus any particular term can be found using $C(n,r)a^{n-r}b^r$C(n,r)an−rbr.
Worked Examples
QUESTION 1
Consider the binomial expansion of $\left(x+\frac{1}{x}\right)^n$(x+1x)n.
Given that the powers of $x$x in the expansion are in decreasing order, write an expression for the $\left(k+1\right)$(k+1)th term.
Fully simplify your answer.
What must be the value of $n$n so that the $8$8th term of the expansion is a constant?
QUESTION 2
Answer the following.
Expand and simplify $\left(1+y\right)^6$(1+y)6
Now expand and simplify $\left(1-y\right)^6$(1−y)6.
Hence, or otherwise, simplify $\left(1+y\right)^6+\left(1-y\right)^6$(1+y)6+(1−y)6
Hence simplify $\left(1+\sqrt{y+2}\right)^6+\left(1-\sqrt{y+2}\right)^6$(1+√y+2)6+(1−√y+2)6.