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:
$(a+b)^n$(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
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?
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.