Question 1

The Olympiad task For the 10th grade if you do not know how to solve it immediately sayCorrect answer 2

The Olympiad task For the 10th grade if you do not know how to solve it immediately-example-1

Question 2

Given:-

To find:-

$s_n=f(1)+f((1)/(2))+f((1)/(3))+\cdots+f((1)/(n))$

Now we simplify above equation to solve it,

$\begin{gathered} f(1)+f((1)/(2))+f((1)/(3))+\cdots+f((1)/(n))=f(1)+f((2-1)/(2))+f((3-2)/(3))+\cdots+f((n+1-n)/(n)) \\ f(1)+f((1)/(2))+f((1)/(3))+\cdots+f((1)/(n))=f(1)+f(1-(1)/(2))+f(1-(2)/(3))+\ldots+f(1-((n-1))/(n)) \\ f(1)+f((1)/(2))+f((1)/(3))+\cdots+f((1)/(n))=f(1)+1-f((1)/(2))+1-f((2)/(3))+\cdots+1-f((n-1)/(n)) \\ f(1)+f((1)/(2))+f((1)/(3))+\cdots+f((1)/(n))=f(1)-f((1)/(2))-f((2)/(3))-\cdots-f((n-1)/(n))+1(n-1) \\ f(1)+f((1)/(2))+f((1)/(3))+\cdots+f((1)/(n))=f(1)-f((1)/(1+1))-f((2)/(2+1))-\cdots-f((n-1)/((n-1)+1))+1(n-1) \end{gathered}$

By furthur simplification we get,

$\begin{gathered} f(1)-f((1)/(1+1))-f((2)/(2+1))-\cdots-f((n-1)/((n-1)+1))+1(n-1)=f(1)-(1)/(2)f(1)-(1)/(2)f(2)-\cdots-(1)/(2)f(n-1)+1(n-1)_{} \\ \text{ =}-(1)/(2)f(1)-(1)/(2)f(2)-\cdots-(1)/(2)f(n-1)+1(n-1) \\ \text{ =-}(1)/(2)\lbrack f(1)+f(2)+...+f(n-1)\rbrack+1(n-1) \end{gathered}$

So the requiired answer is,

$\text{-}(1)/(2)\lbrack f(1)+f(2)+...+f(n-1)\rbrack+1(n-1)$

0 Comments

Your comment on this question:

Your answer

1 Answer

0 Comments

Your comment on this answer:

Other Questions