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

Algebra Level 4

S n = i = 1 n a i S n = i = 1 n 1 a i S ( n ) = i = 1 n ( S n a i 1 ) \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 S(1000)-S_{1000}S_{1000}^{'} .


Try the first problem here .
Try more of my originals in this set .


The answer is -1000.

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1 solution

Nihar Mahajan
Jun 9, 2015

S ( n ) = i = 1 n ( S n a i 1 ) S ( n ) = i = 1 n S n a i i = 1 n 1 S ( n ) = i = 1 n S n a i n S ( n ) = S n a 1 + S n a 2 + S n a 3 S n a n n S ( n ) = S n S n n S ( n ) S n S n = n S ( 1000 ) S 1000 S 1000 = 1000 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

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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

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

Adarsh Kumar - 6 years ago

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