1 से N तक Natural Numbers का Sum — code और dry run

Problem statement

Approach 1: simple loop

Loop from 1 to N and keep adding into an accumulator. The pattern is the same one behind every running total in every program you will write.

n = 5
total = 0

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

print(total)
# Output: 15

Remember that range(1, n + 1) is needed because the stop value is exclusive — range(1, n) would stop one number early.

Approach 2: reusable function

Gauss's formula does it in one step: n × (n + 1) / 2. Wrap it in a function and use integer division // so the answer stays an integer.

def sum_to(n):
    return n * (n + 1) // 2

print(sum_to(5))       # 15
print(sum_to(100))     # 5050
print(sum_to(0))       # 0

The formula is O(1) — one multiplication whether N is 5 or 5 billion. The loop is O(N). Same answer, wildly different cost.

Approach 3: Pythonic alternative

Python's built-in sum() adds anything iterable, and range is iterable — so the loop collapses into one clear line.

n = 5
print(sum(range(1, n + 1)))   # 15

This is the idiomatic middle ground: no manual accumulator to get wrong, and the intent — "sum of the numbers 1..n" — reads off the page.

Dry run

Common errors

• range(1, n) — आखिरी number N ही छूट जाता है।
• Formula में / की जगह // — वरना जवाब float (15.0) बनेगा।
• Accumulator को 1 से शुरू करना — हर total में एक phantom number जुड़ जाएगा।
• Formula में parentheses न लगाना — precedence की वजह से गलत जवाब।
• N = 0 का case भूलना — loop चलेगा ही नहीं, पर initial total 0 ही सही जवाब है।

Practice on PyDebug

Concept समझने के बाद इन free problems को browser में solve करें:

• Range boundary practice
• Operator precedence practice
• Variables and assignment practice