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Of two given natural numbers are formed a new one in the following way: the first is added to the second and obtained the third, the sum of the second and third is the fourth, etc. Find the sum of the first six formed in this way, if the fifth number is 7.

I was doing an IB entrance test and this was on it. Can somebody show me how this is supposed to be done in the easiest way possible?

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You've been given $n_1,n_2\in\mathbb{N}$.

Then, $n_3=n_1+n_2$

$n_4=n_3+n_2=2n_2+n_1$

$n_5=n_4+n_3=2n_2+n_1+n_1+n_2=3n_2+2n_1=7$

Since $n_1$ and $n_2$ are natural numbers, $n_1=2$, $n_2=1$.

So finally the sum is $\sum_{i=1}^6 n_i=2+1+3+4+7+11=28$.

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  • $\begingroup$ Exactly how I did it. Then the final answer is 11 am I correct? $\endgroup$ – RiktasMath Jun 19 '18 at 10:34
  • $\begingroup$ The answer is $\sum_{i=1}^6 n_i=2+1+3+4+7+11=28$. $\endgroup$ – Javi Jun 19 '18 at 17:30
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Note that: $$(n_1+n_2)+n_3+n_4+n_5+(n_6)=(n_3)+n_3+n_4+n_5+(n_4+n_5)=\\ 2(n_3+n_4+n_5)=2(n_5+n_5)=4n_5=28.$$

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