Could someone help me finish this proof? Everything makes sense to me up to the last part... I'm not sure how to prove that $\displaystyle 5d_{k} - 6d_{k-1}=2^{k+1} + 3^{k+1}$
Hi!
If we look at the LHS, it is: $\displaystyle 2^{k+1}+3^{k+1} = 2\cdot 2^k + 3 \cdot 3^k$
RHS: $\displaystyle 5d_k -6d_{k-1} = 5(2^k+3^k) -6(2^{k-1}+3^{k-1}) = 5\cdot 2^k +5\cdot 3^k -6\cdot 2^{k-1}-6\cdot 3^{k-1} = $ $\displaystyle 5\cdot 2^k +5\cdot 3^k - 3\cdot 2^k -2 \cdot 3^k = 2\cdot 2^k + 3\cdot 3^k$
And that is equal to the LHS, and so we are done.