Get sum of sequence using recursion

Question

I have this sum:

Obviously, I have to get sum of that depending on what N is. I need to do it in three different ways.

First is for loop:

function lab(n) {
  var S = 0;
  let VS
  if (n == 0) {
    VS = 0;
    return 0;
  }
  if (n == 1) {
    VS = 4;
    return Math.pow(3 / 5, 1);
  } else {
    for (let i = 0; i < n; i++) { //
      S += 1 / n * Math.pow(3 / 5, n);
      t = 4 * n;
    }
    return S;
  }
}

Second one is recursion:

function lab(n) {
  let vs = 0;
  if (n <= 1)
    return 0;
  else {
    vs += 4 * n // vs is how many actions it takes to make this calculation. I’m sure in for loop this is right, but I’m not sure about the recursion approach
    return lab(n - 1) + 1 / n * Math.pow(3 / 5, n)
  }
}

The third way is use recursion with the condition that in order to get S(n) I need to use S(n-1).

I am stuck on this.

Also I get different sums with the same Ns from first and second function.

score 1 · Accepted Answer · 2017-10-15 19:09:17Z

1

I am not sure what you are asking for.

If you are asking for a recursive function then take a look at the following:

function summation(n, sum = 0) {
  if (n <= 0) {
    return sum;
  }
  
  sum += (1/n) * Math.pow(3/5, n);

  return summation(n - 1, sum);
}

console.log(summation(1));
console.log(summation(2));
console.log(summation(3));
console.log(summation(4));
console.log(summation(5));

Another recursive method without passing sum as parameter:

function summation(n) {
  if (n <= 0) {
    return 0;
  }

  return ((1/n) * Math.pow(3/5, n)) + summation(n - 1);
}

console.log(summation(1));
console.log(summation(2));
console.log(summation(3));
console.log(summation(4));
console.log(summation(5));

Also, for the loop method, the following will suffice

function summation(n) {
  var sum = 0;
    
  while (n > 0) {
      sum += (1/n) * Math.pow(3/5, n);
      n -= 1;
  }

  return sum;
}

console.log(summation(1));
console.log(summation(2));
console.log(summation(3));
console.log(summation(4));
console.log(summation(5));

edited Oct 15, 2017 at 19:09

answered Oct 15, 2017 at 18:39

user4447799

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5 Comments

Tautvydas Būda Over a year ago

These two are great. Also could You tell me how does the number of basic actions change when n is increasing or decreasing. I think for loop is just multiply n by 4, because there are 4 actions (1. / 2. * 3. / 4. ^) so vs=4*n. But how about recurtion?Is it still the same? It's also very confusing for me, because teacher asked us to do simple recursion and recursion that in order to get S(n) i need to use s(n-1), i mean you just did it. Is there any way to do this recursion in different way.

user4447799 Over a year ago

"I think for loop is just multiply n by 4, because there are 4 actions (1. / 2. * 3. / 4. ^) so vs=4*n". I didn't understand what you meant by "basic actions"

user4447799 Over a year ago

That is exactly what I didn't understand, what do you mean by "actions"? I am totally confused by the purpose of VS in your loop. BTW is have aded another recursive method in answer

Tautvydas Būda Over a year ago

Sorry, for confusion! I'll try my best. So by saying action i mean add, subtract, divide, pow etc. So for example this : (1/n) * Math.pow(3/5, n) has 4 action s right? first we divide, then multiply then we divide again and then we pow it. action = mathematical action. so for example in loop function if n=2, then total methematical actions = 8

user4447799 Over a year ago

Oh yes I suppose but the actual order will be division, division, the power and then finally the multiplication. While I don't know why you need to keep track of these actions, but just passing it as parameter will be sufficient in the case of recursion which will increase by 4 in each call.

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