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Apparently $6^7 + 6^8 = 7 \cdot 6^7$, but I'm not sure why this pattern works. I also just plugged in two other examples, $3^1 + 3^2 = 4 \cdot 3^1$ and $4^2 + 4^3 = 5 \cdot 4^2$, it also works.

ervx
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maudulus
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3 Answers3

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$$6^7+6^8=6^7+6\cdot6^7=7\cdot6^7$$

Perhaps using parameters and using exponents laws it can be done clearer:

$$a^7+a^8=a^7+a\cdot a^7=(1+a)a^7$$

and now just substitute $\;a=6\;$ ...or whatever you want.

DonAntonio
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You are saying that $n^a+n^{a+1}=n^a+n^{a}\cdot n=n^{a}(1+n)=(n+1)\cdot n^{a}$.

Robert Z
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$$ 6^{7}+6^{8}=6^{7}+6^{7+1}=6^{7}+6^{7}\cdot 6=6^{7}(1+6)=7\cdot 6^{7}. $$

More generally, for any $x$,

$$ x^{n}+x^{n+1}=x^{n}(1+x). $$

ervx
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