Find the sum of : (i) the first 1000 positive integers (ii) the first n positive integers
Step-by-Step Solution
Key Concept: The sum of the first $n$ terms of an arithmetic progression (AP) is given by $S_n = \frac{n}{2}(a + l)$, where $a$ is the first term and $l$ is the last term. For the series of natural numbers, $a = 1$, $d = 1$, and $l = n$; thus $S_n = \frac{n(n+1)}{2}$.
Step 1: Identify the series as an AP with first term $a=1$ and common difference $d=1$.
Step 2: For the first $n$ positive integers, the $n^{th}$ term (last term) is $l = a + (n-1)d = 1 + (n-1)\times1 = n$.
Step 3: Use the sum formula $S_n = \frac{n}{2}(a + l)$.
Step 4: Substitute $a=1$ and $l=n$:
$$S_n = \frac{n}{2}(1 + n) = \frac{n(n+1)}{2}.$$
Step 5: (i) For $n = 1000$,
$$S_{1000} = \frac{1000\times1001}{2} = 500\times1001 = 500500.$$
Step 6: (ii) For a general $n$, the sum is $S_n = \frac{n(n+1)}{2}$.
Thus the required sums are obtained.
Correct Answer: (i) $500500$\n(ii) $\displaystyle \frac{n(n+1)}{2}$