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Suppose we remove some terms (taking at least 1 item and leaving at least 2 items) from the geometric sequence, $$1, k, k^2, \cdots, k^{n}$$ (with $k>2$) and separate the remaining terms into two groups. Prove these two group's sum can never be the same.

I'm not even sure how to start on this since you can't find a closed form for a sum of group. The entire sum is $$\frac{1(1-k^{n+1})}{1-k}$$ So the we prove that $$\frac{\frac{1(1-k^{n+1})}{1-k}-(k^p+\cdots)}{2}$$ cannot be attained by the remaining terms.

Baker5680
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1 Answers1

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Consider a splitting

$$A = k^{a_1} + \dots +k^{a_s},\quad B = k^{b_1} + \dots +k^{b_r},$$ where $0 \leq a_1 < \dots < a_s \leq n$ and the same holds for the $b_i$s. Without loss of generality suppose $b_r > a_s,$ so

$$A \leq 1 + k + k^2 + \dots + k^{a_s} = \frac{1-k^{a_s+1}}{1-k} < k^{a_s+1} \leq k^{b_r} \leq B,$$

so we get $A < B$.

Brian Moehring
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k76u4vkweek547v7
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