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Algebra Difficulty 5.1 AIME, harder Find the answer

7. Calculate: ∑i=120181i+∑j=12018(∑i=j20181i)2\sum_{i=1}^{2018} \frac{1}{i}+\sum_{j=1}^{2018}\left(\sum_{i=j}^{2018} \frac{1}{i}\right)^{2}

The value is

A number or a short expression. Spacing and $ signs are ignored.

Solution

7.4036.

Let Sn=∑i=1n1i+∑j=1n(∑i=jn1i)2S_{n}=\sum_{i=1}^{n} \frac{1}{i}+\sum_{j=1}^{n}\left(\sum_{i=j}^{n} \frac{1}{i}\right)^{2}.
Then Sn+1−SnS_{n+1}-S_{n}
=1n+1+1n+1∑j=1n(2∑i=jn+11i−1n+1)+(1n+1)2=\frac{1}{n+1}+\frac{1}{n+1} \sum_{j=1}^{n}\left(2 \sum_{i=j}^{n+1} \frac{1}{i}-\frac{1}{n+1}\right)+\left(\frac{1}{n+1}\right)^{2}
⇒Sn+1−Sn=2\Rightarrow S_{n+1}-S_{n}=2.
Also, S1=2,Sn=2nS_{1}=2, S_{n}=2 n, so S2018=4036S_{2018}=4036.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.