Sum the n odd terms ,using Loop for ,without using list, allowed functions:input,int,print and range

Question

n =(int(input("Digite o número de termos:")))

for i in range(n+n):
    if i%2>0:
      print(i)     

#Não sei como prosseguir depois,como pego esses valores e somo eles 

Answer 1

A detail: in the title it is said that you want to add the "n odd terms", and in the code is the message "Digite o número de termos". Does that mean that n is the amount of odd numbers, right?

For example, if n for 5, Do you want to add up all the odd numbers from 1 to 5 (i.e., 1 + 3 + 5), or do you want to add up the first 5 odd numbers (i.e., 1 + 3 + 5 + 7 + 9)? The other answers understood that it is the first option...

Anyway, let’s see a solution for each:

---

Add the odd ones from 1 to n

If so, there are several options. It can be with a simple loop:

n = int(input("Digite o número de termos:"))
soma = 0
for i in range(1, n + 1, 2):
    soma += i
print(soma) # 9 (1 + 3 + 5)

Note that I use the third parameter of range, which is the step, that is, the increment that is added to each element to generate the next one. If the range begins in 1 and I just want the odd numbers, I can jump from 2 to 2. No need none to advance 1 by 1 and go testing whether the number is even or odd (unless it is an exercise that "demand" do this).

Also note that the range ends in n + 1, because the last number is not included (ie if I just put n, own n would not be added).

Another way to do it is to use sum, along with a Generator Expression, much more succinct and pythonic:

soma = sum(i for i in range(1, n + 1, 2))
print(soma)

Although sum receives an iterable parameter, and ranges are eternal, so you can simplify even more:

soma = sum(range(1, n + 1, 2))
print(soma)

---

Add up the n first odd numbers

If when n is, for example, equal to 5, and the goal is to add up the first 5 odd numbers (i.e., 1 + 3 + 5 + 7 + 9), there it is like this:

soma = 0
impar = 1
for i in range(n):
    soma += impar
    impar += 2
print(soma) # 25 (1 + 3 + 5 + 7 + 9)

The range(n) serves to control the amount of odd numbers, and the odd number starts at 1 and is incremented by 2 in 2.

The for above can also be more succinct:

soma, impar = 0, 1
for i in range(n):
    soma, impar = soma + impar, impar + 2

Although the nth odd number is equal to 2n - 1, then it would be enough to use similar solutions to the first case, but now using a range until 2n:

soma = 0
for i in range(1, 2 * n, 2):
    soma += i
print(soma)

# ou
soma = sum(range(1, 2 * n, 2))
print(soma)

---

But in fact this problem boils down to the sum of a PA (arithmetic progression), and in that case there is a ready-made formula for the sum of the first N terms:

(a1 + an) * n / 2

• a1 is the first term (in our case it is 1)
• an is the umpteenth term, which in turn is equal to a1 + (n - 1) * r (and r is the reason, that in our case is 2 - the difference between each term of PA)

So just apply the formula:

n = int(input("Digite o número de termos:"))

soma = (1 + 1 + (n - 1) * 2) * n / 2

---

But note that if you simplify the above formula with these specific values (a1=1 and r=2), will come to the conclusion that the sum of n first odd numbers is equal to n squared, then a more succinct way is still:

soma = n * n

# ou
soma = n ** 2

---

Note: the AP formula can also be applied for the first case (adding odd numbers from 1 to n). But in that case n is the nth term, and the quantity of terms shall be calculated:

# calcule a quantidade de termos
qtd = (1 + n) // 2

# use a fórmula da soma da PA
soma = (1 + n) * qtd / 2

# ou, como é a soma dos primeiros "qtd" ímpares, use a "fórmula simplificada"
soma = qtd ** 2

---

But if it’s an exercise that "requires" using a for 1 in 1 and also the operator %, then the solution is:

# soma dos ímpares de 1 a n
soma = 0
for i in range(1, n + 1):
    if i % 2 != 0:
        soma += i
print(soma)

# soma dos n primeiros ímpares
soma = 0
for i in range(1, 2 * n):
    if i % 2 != 0:
        soma += i
print(soma)

Answer 2

There are some problems, like your range that is increasing (doubling) the amount of terms:

range(n+n)

To get the values, just access the loop variable, you already do this, but just print the contents of it:

print(i)

---

To add the terms, you will need another variable, we will create it with the value of zero and call it sum, to be very evident your purpose:

soma = 0

In the for, we’ll use it from 1 to the end:

for i in range(1, n+1):

Now inside the for, let’s add the value in the variable soma whenever it is odd;

if i%2>0:
  soma += i

And finally, we only print the variable value soma:

print(soma)

---

Putting it all together, the code will look like this:

n = int(input("Digite o número de termos:"))
soma = 0

for i in range(1, n+1):
  if i%2 > 0:
    soma += i

print(soma)

---

See online: https://repl.it/repls/RashFortunateLevels

Answer 3

Here’s the solution. I don’t understand what you’re adding up to n + n?

n =(int(input("Digite o número de termos:")))

result = 0

for i in range(n):
  if not i % 2 == 0:
    result += i


print(result)

Answer 4

In this matter you have to pay attention to two things.

First thing:

Define an accumulating variable (variable that will store the calculation performed in each iteration of "for"). Variable that I will call "x".

Second thing:

Structure the for correctly.

If you intend to use the repeat loop "for", iterating over a range, you have to set the limits of this range. In this case, the for would look the way...

for c in range(1, n + 1):

...where "1" would be the lower limit and "n + 1" the upper limit.

With these two situations in mind, I developed the following algorithm below.

# Capturando e tratando o valor inserido no input.
while True:
    try:
        n = int(input('Digite o número de termos: '))
        if n <= 0:
            print('\033[31mValor INVÁLIDO! Digite apenas inteiros maiores que "0"!\033[m')
        else: 
            break
    except:
        print('\033[31mValor INVÁLIDO! Digite apenas números inteiros!\033[m')

# Realizando os cálculos e exibindo o resultado:
x = 0
for c in range(1, n + 1):
    if c % 2 != 0:
        x += c
print(f'O resultado é: {x}')

Note how the code works on Repl.it

Also note that this algorithm performs a treatment of the values captured by the input.