What is the required recursive function(s) in Scheme programming language to compute the following series?

Question

what is the required recursive function(s) in Scheme programming language to compute the following series?? Explanation needed

1^2/2^1 + 3^4/4^3 + 5^6/6^5 +

Answer 1

So, well, what does each term look like? It's n^(n+1)/(n+1)^n. And you want to stop when you reach 10 (so if n > 10, stop). So write a function of a single argument, n, which either:

• returns 0 if n > 10;
• adds n^(n+1)/(n+1)^n to the result of calling itself on n + 2.

Then this function with argument 1 will compute what you want. Going backwards may be easier:

• return 0 if n < 1;
• add n^(n+1)/(n+1)^n to the result of calling itself on n - 2;

then the function with argument 10 is what you want.

---

Or you could do this which is more entertaining:

(define s
  (λ (l)
    ((λ (c i a)
       (if (> i l)
           a
           (c c
              (+ i 2)
              (+ a (/ (expt i (+ i 1))
                      (expt (+ i 1) i))))))
     (λ (c i a)
       (if (> i l)
           a
           (c c
              (+ i 2)
              (+ a (/ (expt i (+ i 1))
                      (expt (+ i 1) i))))))
     1 0)))

But I don't recommend it.

Answer 2

//power function
(define (power a b)
     (if (zero? b) //base case
      1    
     (* a (power a (- b 1))))) //or return power of a,b

// sum function for series

    (define (sum n)
     (if (< n 3) //base case 
         0.5
       (+ (/  (power (- n 1) n) (power n (- n 1))) (sum (- n 2 ))  ))) //recursion call

>(sum 10) // call sum function here .