a) If x = y1! + y^22! + y^33! + , show that y = x x^22 + x^33 [2] b) Using mathematical induction, prove that : n! 2^n for n 4 [2] c) Use Cramer's rule to solve the system of linear equations. 2x +…

a) Show that $y=x-\frac{x^2}{2}+\frac{x^3}{3}-\dots$ given $x=\frac{y}{1!}+\frac{y^2}{2!}+\frac{y^3}{3!}+\dots$

Step 1: Recognize the series for $x$ The given series for $x$ is: $$x=\sum _{k=1}^{\infty }\frac{y^k}{k!}=\frac{y}{1!}+\frac{y^2}{2!}+\frac{y^3}{3!}+\dots$$ This resembles the Taylor series expansion of $e^y$, but without the $k=0$ term: $$e^y=\sum _{k=0}^{\infty }\frac{y^k}{k!}=1+\frac{y}{1!}+\frac{y^2}{2!}+\frac{y^3}{3!}+\dots$$ Thus, we can write: $$x=e^y-1$$

Step 2: Solve for $y$ in terms of $x$ We have: $$x+1=e^y$$ Taking the natural logarithm of both sides: $$\ln (x+1)=y$$ Now, expand $\ln (x+1)$ as a Taylor series around $x=0$: $$\ln (1+x)=\sum _{k=1}^{\infty }(-1{)}^{k+1}\frac{x^k}{k}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots$$ Thus: $$y=\ln (x+1)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots$$

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b) Prove by mathematical induction that $n!\ge 2^n$ for $n\ge 4$

Base Case (n = 4): Calculate $4!$ and $2^4$: $$4!=24,2^4=16$$ Since $24\ge 16$, the base case holds.

Inductive Step: Assume the statement is true for some $k\ge 4$, i.e., $$k!\ge 2^k$$ We must show that $(k+1)!\ge 2^{k+1}$.

Starting from the inductive hypothesis: $$(k+1)!=(k+1)\cdot k!$$ By the inductive hypothesis, $k!\ge 2^k$, so: $$(k+1)!\ge (k+1)\cdot 2^k$$ Since $k\ge 4$, $k+1\ge 5$, and $5\cdot 2^k\ge 2^{k+1}$ because: $$5\cdot 2^k\ge 2^{k+1}⟹5\ge 2$$ which is true. Thus: $$(k+1)!\ge 2^{k+1}$$

Conclusion: By the principle of mathematical induction, $n!\ge 2^n$ for all $n\ge 4$.

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c) Solve the system of linear equations using Cramer's rule:

$$\{\begin{array}{l}2x+y=4\\ 3y-z=3\\ 2x+3z=11\end{array}$$

Step 1: Write the system in matrix form $AX=B$ $$A=\left[\begin{array}{ccc}2& 1& 0\\ 0& 3& -1\\ 2& 0& 3\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],B=\left[\begin{array}{c}4\\ 3\\ 11\end{array}\right]$$

Step 2: Compute the determinant of $A$ ($\Delta$) $$\Delta =\left|\begin{array}{ccc}2& 1& 0\\ 0& 3& -1\\ 2& 0& 3\end{array}\right|=2\cdot \left|\begin{array}{cc}3& -1\\ 0& 3\end{array}\right|-1\cdot \left|\begin{array}{cc}0& -1\\ 2& 3\end{array}\right|+0\cdot \left|\begin{array}{cc}0& 3\\ 2& 0\end{array}\right|$$ $$=2\cdot (9-0)-1\cdot (0+2)+0=18-2=16$$

Step 3: Compute ${\Delta }_x$ (replace first column of $A$ with $B$) $${\Delta }_x=\left|\begin{array}{ccc}4& 1& 0\\ 3& 3& -1\\ 11& 0& 3\end{array}\right|=4\cdot \left|\begin{array}{cc}3& -1\\ 0& 3\end{array}\right|-1\cdot \left|\begin{array}{cc}3& -1\\ 11& 3\end{array}\right|+0\cdot \left|\begin{array}{cc}3& 3\\ 11& 0\end{array}\right|$$ $$=4\cdot (9-0)-1\cdot (9+11)=36-20=16$$

Step 4: Compute ${\Delta }_y$ (replace second column of $A$ with $B$) $${\Delta }_y=\left|\begin{array}{ccc}2& 4& 0\\ 0& 3& -1\\ 2& 11& 3\end{array}\right|=2\cdot \left|\begin{array}{cc}3& -1\\ 11& 3\end{array}\right|-4\cdot \left|\begin{array}{cc}0& -1\\ 2& 3\end{array}\right|+0\cdot \left|\begin{array}{cc}0& 3\\ 2& 11\end{array}\right|$$ $$=2\cdot (9+11)-4\cdot (0+2)=40-8=32$$

Step 5: Compute ${\Delta }_z$ (replace third column of $A$ with $B$) $${\Delta }_z=\left|\begin{array}{ccc}2& 1& 4\\ 0& 3& 3\\ 2& 0& 11\end{array}\right|=2\cdot \left|\begin{array}{cc}3& 3\\ 0& 11\end{array}\right|-1\cdot \left|\begin{array}{cc}0& 3\\ 2& 11\end{array}\right|+4\cdot \left|\begin{array}{cc}0& 3\\ 2& 0\end{array}\right|$$ $$=2\cdot (33-0)-1\cdot (0-6)+4\cdot (0-6)=66+6-24=48$$

Step 6: Solve for $x,y,z$ using Cramer's rule $$x=\frac{{\Delta }_x}{\Delta }=\frac{16}{16}=1$$ $$y=\frac{{\Delta }_y}{\Delta }=\frac{32}{16}=2$$ $$z=\frac{{\Delta }_z}{\Delta }=\frac{48}{16}=3$$

Final solution: $$\boxed{x=1,y=2,z=3}$$