a) Sum to n terms of the series 1^2.1 + 2^2.3 + 3^2.5 + [2] b) Solve the following system of linear equations by using matrix inversion method. x + 2y = 5 , 3x y = 2 . [3]

a) Sum to n terms of the series $1^2.1+2^2.3+3^2.5+\dots$

The given series is: $$S=1^2\cdot 1+2^2\cdot 3+3^2\cdot 5+\cdots +n^2\cdot (2n-1)$$

We need to find the sum of the first $n$ terms of this series.

Step 1: Identify the general term

The $k$-th term of the series is: $$T_k=k^2\cdot (2k-1)$$ $$T_k=2k^3-k^2$$

Step 2: Find the sum of the series

The sum of the first $n$ terms is: $$S_n=\sum _{k=1}^n{T}_k=\sum _{k=1}^n(2k^3-k^2)$$ $$S_n=2\sum _{k=1}^n{k}^3-\sum _{k=1}^n{k}^2$$

We know the following standard formulas: $$\sum _{k=1}^n{k}^3={\left(\frac{n(n+1)}{2}\right)}^2$$ $$\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$$

Substitute these into the expression for $S_n$: $$S_n=2{\left(\frac{n(n+1)}{2}\right)}^2-\frac{n(n+1)(2n+1)}{6}$$ $$S_n=2\cdot \frac{n^2(n+1{)}^2}{4}-\frac{n(n+1)(2n+1)}{6}$$ $$S_n=\frac{n^2(n+1{)}^2}{2}-\frac{n(n+1)(2n+1)}{6}$$

Factor out $\frac{n(n+1)}{6}$: $$S_n=\frac{n(n+1)}{6}\left[3n(n+1)-(2n+1)\right]$$ $$S_n=\frac{n(n+1)}{6}\left[3n^2+3n-2n-1\right]$$ $$S_n=\frac{n(n+1)}{6}\left[3n^2+n-1\right]$$

Thus, the sum to $n$ terms of the series is: $$\boxed{S_n=\frac{n(n+1)(3n^2+n-1)}{6}}$$

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b) Solve the following system of linear equations by using matrix inversion method.

Given system of equations: $$\{\begin{array}{l}x+2y=5\\ 3x-y=2\end{array}$$

Step 1: Write the system in matrix form $AX=B$

$$\left(\begin{array}{cc}1& 2\\ 3& -1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}5\\ 2\end{array}\right)$$

Step 2: Find the inverse of matrix $A$

The inverse of a $2\times 2$ matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ is given by: $$A^{-1}=\frac{1}{ad-bc}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$$

For our matrix: $$A=\left(\begin{array}{cc}1& 2\\ 3& -1\end{array}\right)$$ $$\text{Determinant}=(1)(-1)-(2)(3)=-1-6=-7$$

Thus, $$A^{-1}=\frac{1}{-7}\left(\begin{array}{cc}-1& -2\\ -3& 1\end{array}\right)$$ $$A^{-1}=\left(\begin{array}{cc}\frac{1}{7}& \frac{2}{7}\\ \frac{3}{7}& -\frac{1}{7}\end{array}\right)$$

Step 3: Multiply $A^{-1}$ with $B$ to find $X$

$$X=A^{-1}B$$ $$\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}\frac{1}{7}& \frac{2}{7}\\ \frac{3}{7}& -\frac{1}{7}\end{array}\right)\left(\begin{array}{c}5\\ 2\end{array}\right)$$

Calculate each component: $$x=\frac{1}{7}\cdot 5+\frac{2}{7}\cdot 2=\frac{5}{7}+\frac{4}{7}=\frac{9}{7}$$ $$y=\frac{3}{7}\cdot 5+\left(-\frac{1}{7}\right)\cdot 2=\frac{15}{7}-\frac{2}{7}=\frac{13}{7}$$

Thus, the solution is: $$\boxed{x=\frac{9}{7},y=\frac{13}{7}}$$