Peace Operator

note that: $a_1☮a_2☮a_3☮....☮a_n=s_1+s_2+s_3+....+s_n$ where $s_n$ is the nth symmetric sum of $a_1,a_2,a_3....,,a_n$ . it can be written as $s_1+s_2+s_3+....+s_n+1-1=(a_1+1)(a_2+1)....(a_n+1)-1$ put the fraction, and we get $(\frac{1}{2}+1)(\frac{1}{3}+1)....(\frac{1}{2015}+1)-1$ $\frac{3}{2}\times \frac{4}{3}....\times \frac{2016}{2015} -1$ telescope $\frac{2016}{2}-1=1008-1=\boxed{1007}$

Let n denote the denominator of the last fraction in a sequence like that given in the problem, and S(n) the value of the evaluated sequence at n. Computing the first several iterations of the operation leads us to this pattern:

• S(2) = .5
• S(3) = 1
• S(4) = 1.5
• S(5) = 2

It is easy to recognize this pattern as the linear function S(n) = .5n-.5. Evaluating S(n) at n=2015 will lead to the value 1007.

However, this is unsatisfying, because we want confirmation the pattern will continue. We can use induction for this. Let P(n) be the statement that "For the integer n, S(n) = .5n-.5"

It is simple to confirm that S(3) = .5(3)-.5 = 1 using the definition provided; this is the base case for induction.

Now suppose that P(k) is true for some k; that is, S(k)=.5k-.5 for some integer k in the sequence.

Using the definition of our sequence: $S(k+1) = S(k) + \frac{1}{k+1} + S(k) \frac{1}{k+1}$

Using the information from our induction hypothesis and simplifying by combining fractions leads us to:

$S(k+1) = .5k-.5 + \frac{1}{k+1} + (.5k-.5) \frac{1}{k+1}$

$S(k+1) = .5(k-1)\frac{k+1}{k+1} + \frac{1}{k+1} + \frac{.5k-.5}{k+1}$

$S(k+1) = \frac{.5 k^2 + .5k}{k+1}$

$S(k+1) = .5k$

This appears not to satisfy our induction process, but shuffling a few numbers around creatively reveals that:

$S(k+1) = .5k + .5 - .5$

$S(k+1) = .5(k+1) - .5$

And therefore P(k+1) is true.

Thus, by the principle of mathematical induction, the linear pattern that we see is guaranteed to continue indefinitely.

Thats easy its like (1+a)(1+b)(1+c)-------