Question

1. write down the expansion of the following using pascals triangle

(c + 3)^4

1. find the coefficient of x^3 in the expansion of the following using pascals triangle

(3x + 1)^4

1. expand fully the following

(1 + 3x)(1 - 2x)^4

1. the coefficient of x^2 in the expansion (d - x)^3 is -9. find the value of d.
2. the coefficient of x^2 in the expansion (2 + ax)^3 is 6. find the values of a.
3. find the value of the following

a) 14!/11!
b) 8!/((8 - 5)!5!)
c) (6 choose 5)

1. find the value of n given that ^nc_2 = 171.

Explanation:

Step1: Recall Pascal's triangle for power 4

The coefficients for $(a + b)^4$ from Pascal's triangle are 1, 4, 6, 4, 1. So, $(c + 3)^4=1\times c^{4}\times3^{0}+4\times c^{3}\times3^{1}+6\times c^{2}\times3^{2}+4\times c^{1}\times3^{3}+1\times c^{0}\times3^{4}$.

Step2: Calculate each term

$=c^{4}+12c^{3}+54c^{2}+108c + 81$

Step3: For $(3x + 1)^4$, use Pascal's triangle coefficients

Coefficients are 1, 4, 6, 4, 1. The general term is $T_{r + 1}=C(n,r)a^{n - r}b^{r}$. Here $n = 4$, $a=3x$, $b = 1$. For the $x^{3}$ term, $n-r=3$, so $r = 1$. Then $T_{2}=4\times(3x)^{3}\times1^{1}=4\times27x^{3}=108x^{3}$. The coefficient of $x^{3}$ is 108.

Step4: Expand $(1 - 2x)^4$ using Pascal's triangle

Coefficients are 1, 4, 6, 4, 1. So $(1 - 2x)^4=1-8x + 24x^{2}-32x^{3}+16x^{4}$. Then $(1 + 3x)(1 - 2x)^4=(1 + 3x)(1-8x + 24x^{2}-32x^{3}+16x^{4})=1-8x + 24x^{2}-32x^{3}+16x^{4}+3x-24x^{2}+72x^{3}-96x^{4}+48x^{5}=1-5x + 40x^{3}-80x^{4}+48x^{5}$

Step5: Expand $(d - x)^3$ using Pascal's triangle

Coefficients are 1, 3, 3, 1. So $(d - x)^3=d^{3}-3d^{2}x + 3dx^{2}-x^{3}$. The coefficient of $x^{2}$ is $3d$. Given $3d=-9$, then $d=-3$.

Step6: Expand $(2+ax)^3$ using Pascal's triangle

Coefficients are 1, 3, 3, 1. So $(2 + ax)^3=8 + 12ax+6a^{2}x^{2}+a^{3}x^{3}$. The coefficient of $x^{2}$ is $6a^{2}$. Given $6a^{2}=6$, then $a^{2}=1$, so $a=\pm1$.

Step7: Calculate factorial - related values

a) $\frac{14!}{11!}=\frac{14\times13\times12\times11!}{11!}=14\times13\times12 = 2184$.
b) $\frac{8!}{(8 - 5)!5!}=\frac{8!}{3!5!}=\binom{8}{5}=\frac{8\times7\times6\times5!}{3\times2\times1\times5!}=56$.
c) $\binom{6}{5}=\frac{6!}{5!(6 - 5)!}=\frac{6!}{5!1!}=6$

Step8: Solve for $n$ in $\binom{n}{2}=171$

We know $\binom{n}{2}=\frac{n!}{2!(n - 2)!}=\frac{n(n - 1)}{2}=171$. Then $n(n - 1)=342$, $n^{2}-n - 342=0$. Factoring gives $(n - 19)(n+18)=0$. So $n = 19$ or $n=-18$. Since $n>0$ in combination context, $n = 19$.

Answer:

1. $(c + 3)^4=c^{4}+12c^{3}+54c^{2}+108c + 81$
2. 108
3. $(1 + 3x)(1 - 2x)^4=1-5x + 40x^{3}-80x^{4}+48x^{5}$
4. $d=-3$
5. $a=\pm1$
6. a) 2184; b) 56; c) 6
7. $n = 19$