a) If the middle term in the expansion ( a2+2)^8 is 1120, find the value of a. [2] b) Using mathematical induction, prove that 1+7+13+19+...+(6n 5)=n(3n 2) [3] c) Solve the following linear…

a) Finding the value of $a$ in the binomial expansion

Given: The middle term in the expansion of ${\left(\frac{a}{2}+2\right)}^8$ is 1120.

Step 1: Identify the middle term For the expansion of $(x+y{)}^n$, the number of terms is $n+1$. Here, $n=8$, so the number of terms is $9$. The middle term is the $\frac{9+1}{2}=5^{th}$ term.

Step 2: Write the general term The general term in the expansion of ${\left(\frac{a}{2}+2\right)}^8$ is given by: $$T_{k+1}=\left(\genfrac{}{}{0ex}{}{8}{k}\right){\left(\frac{a}{2}\right)}^{8-k}(2{)}^k$$ For the $5^{th}$ term ($k=4$): $$T_5=\left(\genfrac{}{}{0ex}{}{8}{4}\right){\left(\frac{a}{2}\right)}^4(2{)}^4$$

Step 3: Simplify and set equal to 1120 $$\left(\genfrac{}{}{0ex}{}{8}{4}\right){\left(\frac{a}{2}\right)}^4(2{)}^4=1120$$ $$70\cdot \frac{a^4}{16}\cdot 16=1120$$ $$70\cdot a^4=1120$$ $$a^4=\frac{1120}{70}=16$$ $$a=\pm 2$$

Final Answer: The value of $a$ is $\boxed{\pm 2}$.

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b) Proof by Mathematical Induction

Statement to prove: $$1+7+13+19+\cdots +(6n-5)=n(3n-2)$$

Step 1: Base case ($n=1$) For $n=1$, the left-hand side (LHS) is: $$6(1)-5=1$$ The right-hand side (RHS) is: $$1(3(1)-2)=1$$ Since LHS = RHS, the base case holds.

Step 2: Inductive hypothesis Assume the statement is true for $n=k$: $$1+7+13+\cdots +(6k-5)=k(3k-2)$$

Step 3: Inductive step ($n=k+1$) For $n=k+1$, the LHS becomes: $$1+7+13+\cdots +(6k-5)+[6(k+1)-5]$$ Using the inductive hypothesis: $$=k(3k-2)+(6k+6-5)$$ $$=3k^2-2k+6k+1$$ $$=3k^2+4k+1$$

Now, the RHS for $n=k+1$ is: $$(k+1)(3(k+1)-2)=(k+1)(3k+3-2)=(k+1)(3k+1)$$ $$=3k^2+k+3k+1=3k^2+4k+1$$

Since LHS = RHS, the statement holds for $n=k+1$.

Conclusion: By the principle of mathematical induction, the statement is true for all positive integers $n$.

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c) Solving the system of linear equations using the matrix method

Given system: $$\{\begin{array}{l}7x-2y=18\\ 3x+7z=33\\ x+y+z=12\end{array}$$

Step 1: Write the augmented matrix The system can be represented as: $$\left[\begin{array}{ccccc}7& -2& 0& |& 18\\ 3& 0& 7& |& 33\\ 1& 1& 1& |& 12\end{array}\right]$$

Step 2: Perform row operations to obtain the row-echelon form

Row 1 ↔ Row 3 (to make the leading coefficient 1 in the first row): $$\left[\begin{array}{ccccc}1& 1& 1& |& 12\\ 3& 0& 7& |& 33\\ 7& -2& 0& |& 18\end{array}\right]$$

Row 2 → Row 2 - 3 × Row 1: $$\left[\begin{array}{ccccc}1& 1& 1& |& 12\\ 0& -3& 4& |& -3\\ 7& -2& 0& |& 18\end{array}\right]$$

Row 3 → Row 3 - 7 × Row 1: $$\left[\begin{array}{ccccc}1& 1& 1& |& 12\\ 0& -3& 4& |& -3\\ 0& -9& -7& |& -66\end{array}\right]$$

Row 2 → Row 2 × (-1/3): $$\left[\begin{array}{ccccc}1& 1& 1& |& 12\\ 0& 1& -\frac{4}{3}& |& 1\\ 0& -9& -7& |& -66\end{array}\right]$$

Row 3 → Row 3 + 9 × Row 2: $$\left[\begin{array}{ccccc}1& 1& 1& |& 12\\ 0& 1& -\frac{4}{3}& |& 1\\ 0& 0& -19& |& -57\end{array}\right]$$

Row 3 → Row 3 × (-1/19): $$\left[\begin{array}{ccccc}1& 1& 1& |& 12\\ 0& 1& -\frac{4}{3}& |& 1\\ 0& 0& 1& |& 3\end{array}\right]$$

Step 3: Back substitution

From Row 3: $$z=3$$

From Row 2: $$y-\frac{4}{3}z=1⟹y-\frac{4}{3}(3)=1⟹y-4=1⟹y=5$$

From Row 1: $$x+y+z=12⟹x+5+3=12⟹x=4$$

Final Answer: The solution is $x=\boxed{4}$, $y=\boxed{5}$, and $z=\boxed{3}$.