1000 followers problem - (2) (Easier than the first)

Algebra Level 4

$\begin{aligned} S_{n}&=&\displaystyle\sum_{i=1}^{n} a_i \\ S_{n}^{'}&=&\displaystyle \sum_{i=1}^{n} \dfrac{1}{a_i} \\ S(n)&=&\displaystyle\sum_{i=1}^{n} (S_{n}^{'}a_i - 1) \end{aligned}$

I define three kinds of very similar functions as above.

Find the value of $S(1000)-S_{1000}S_{1000}^{'}$ .

Try the first problem here .

Try more of my originals in this set .

The answer is -1000.

1 solution

Nihar Mahajan
Jun 9, 2015

$S(n)=\displaystyle\sum_{i=1}^{n} (S_{n}^{'}a_i - 1) \\ \Rightarrow S(n)=\displaystyle\sum_{i=1}^{n} S_{n}^{'}a_{i} - \displaystyle\sum_{i=1}^{n} 1 \\ \Rightarrow S(n)=\displaystyle\sum_{i=1}^{n} S_{n}^{'}a_{i} - n \\ \Rightarrow S(n)= S_{n}^{'}a_{1} + S_{n}^{'}a_{2}+ S_{n}^{'}a_{3} \dots S_{n}^{'}a_{n} - n \\ \Rightarrow S(n)= S_nS_{n}^{'}-n \\ \Rightarrow S(n)-S_nS_{n}^{'}=-n \\ \Rightarrow S(1000)-S_{1000}S_{1000}^{'} = \boxed{-1000}$

You know,at first i was intimidated by this problem but after solving it it was like 2+3!

Adarsh Kumar - 6 years ago

I know. Many people are intimidated by this problem but only few people like you are able to see the simplicity of the problem. :3

Nihar Mahajan - 6 years ago

thanx!BTW when is your school reopening?reply on gmail!