Help finishing a proof

**centenial** · October 21st 2009, 08:21 PM

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 $5d_{k} - 6d_{k-1}=2^{k+1} + 3^{k+1}$

**Defunkt** · October 21st 2009, 09:28 PM

Hi!

If we look at the LHS, it is: $2^{k+1}+3^{k+1} = 2\cdot 2^k + 3 \cdot 3^k$

RHS: $5d_k -6d_{k-1} = 5(2^k+3^k) -6(2^{k-1}+3^{k-1}) = 5\cdot 2^k +5\cdot 3^k -6\cdot 2^{k-1}-6\cdot 3^{k-1} =$ $5\cdot 2^k +5\cdot 3^k - 3\cdot 2^k -2 \cdot 3^k = 2\cdot 2^k + 3\cdot 3^k$

And that is equal to the LHS, and so we are done.

**centenial** · October 21st 2009, 09:33 PM

I spent three hours banging my head on that problem tonight, I really appreciate your response. I understand how it works now. Thanks so much!

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