An algebra problem by mridul jain

My solution is very similar to the one of Yugesh Kothari, but this one is an attempt to explain it in proper LaTeX and with some additional details. We consider the sequence $s_n=ax^n+by^n$ . Then we have that $s_1=3,$ $s_2=7,$ $s_3=16,$ and $s_4=42.$ By direct calculation we get that $s_{n+1}=\alpha\:s_n-\beta\: s_{n-1},\:\:\:\:\:\:\:(*)$ where $\alpha=x+y$ and $\beta=xy.$

Making $n=2,$ in $(*),$ we obtain the equation $16=7 \alpha-3\beta,$ and making $n=3$ again in $(*),$ we get $42=16 \alpha- 7\beta.$ Solving the system formed by these two equations we get $\alpha=-14$ and $\beta=-38.$ Now making $n=4$ in $(*)$ for the third time, we get $s_5=\alpha\:s_4-\beta\: s_3=-14(42)-(-38)(16)=20.$ So $ax^5+by^5=20.$

let x+y=p, xy=q. We know, (ax^2 + by^2)(x+y)=(ax^3+by^3)+xy(ax+by). => 7p=16+3q. Applying same process on (ax^3+by^3) 16p=42+7q.

Solving, we get p=-14,q=-38. Applying same process on (ax^4+by^4) we get the answer as 20.

Considering the fact that $ax^n +by^n$ is a solution to a linear homogeneous recurrence relation of degree two.
So,we find the recurrence relation $t_n = \alpha t_{n-1} + \beta t_{n-2}\text{,}$ by finding α,β. $\\$ 3α+7β=16 $\\$ 7α+16β=42 $\\$ Solving eqns we get $α=38 , β=-14 \\ ax^5 +by^5=t_{5} = 16α + 42β = 20$