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Worked solution

Split an infinite GP into equal prefix and tail sums

In an infinite geometric progression with a nonzero first term and common ratio 126\frac1{\sqrt[6]{2}}, the sum of the first nn terms equals the sum of all the remaining terms. Find nn.

[Adapted from SMO 2016 Open Section Round 1 Q2]

CJC · Sequences and Series · Example 28

Answer

n=6n=6.

Compare your reasoning

Express the remaining terms as a tail

Sn=S∞−Sn⟹S∞=2SnS_n=S_\infty-S_n\quad\Longrightarrow\quad S_\infty=2S_n

The tail excludes the first n terms. Since the two halves are equal, the full infinite sum is twice the prefix.

Substitute the two geometric sums

a1−2−1/6=2a[1−(2−1/6)n]1−2−1/6,1=2[1−(2−1/6)n],(2−1/6)n=12.\begin{aligned}\frac{a}{1-2^{-1/6}}&=2\frac{a[1-(2^{-1/6})^n]}{1-2^{-1/6}},\\1&=2[1-(2^{-1/6})^n],\\(2^{-1/6})^n&=\frac12.\end{aligned}

Write r=2^(−1/6). Its modulus is below one, and a≠0 allows cancellation.

Match powers of two

2n/6=21⟹n6=1⟹n=62^{n/6}=2^1\quad\Longrightarrow\quad\frac n6=1\quad\Longrightarrow\quad n=6

The source rewrites the reciprocal power as 2^(n/6)=2^1 and equates exponents.

How did this question go?

Common mistakes

The tail starts with term n+1, so its total is S∞−S_n. The nonzero-first-term condition is needed: an all-zero sequence would satisfy the equality for every n.