Question

Example 14: Find the sum of
(i) the first 1000 positive integers
(ii) the first n positive integers​

Answer 1

To Find:

The sum of,

1. The first 1000 positive integers.
2. The first n positive integers.

We know that:

• Sₙ = {n(a + l)}/2

Where,

• Sₙ = Sum of nth term
• n = Number of terms
• a = First term
• l = Last term

Finding the sum of first 1000 positive integers:

1 + 2 + 3 + 4 + . . . . . . + 1000

We have,

• a = a₁ = 1
• l = n = 1000

S₁₀₀₀ = {1000(1 + 1000)}/2

S₁₀₀₀ = {1000 × 1001}/2

S₁₀₀₀ = 500 × 1001

S₁₀₀₀ = 500500

Hence,

• Sum of first 1000 positive integers = 500500

Finding the sum of first n positive integers:

1 + 2 + 3 + 4 + . . . . . . + n

We have,

• a = a₁ = 1
• l = n = n

Sₙ = {n(1 + n)}/2

Sₙ = {n + n²}/2

Hence,

• Sum of first n positive integers = {n + n²}/2

Answer 2

━━━━━━━━━━━━━━━━━━━━━━━━━━

★ Step by step explanation :-

• Here, we have to find out the sum of first 1000 positive integers and the sum of first n positive integers.

★ Solving for first case :-

• AP = 1 + 2 + 3 . . . . . . + 1000

Therefore,

• n = 1000
• l = 1000
• a = 1

★ Using formula :-

$\qquad\qquad\bf\dag\:\bigg({\sf{\pink{S_{n} = \dfrac{n}{2}\big\lgroup a + l \big \rgroup}}}\bigg)$

★ Putting all known values :-

$\qquad\leadsto\quad\sf S_{1000} = \dfrac{1000}{2}\big\lgroup 1 + 1000\big \rgroup$

$\qquad\leadsto\quad\sf S_{1000} = \dfrac{\cancel{1000}}{\cancel{2}}\big\lgroup 1001\big \rgroup$

$\qquad\leadsto\quad\sf S_{1000} = 500\:\times\:1001$

$\qquad\leadsto\quad{\bf{\purple{S_{1000} = 500500}}}$

$\small\therefore\:{\underline{\sf{Hence,\:sum\:of\:first\:\bf{1000}\:\sf{positive\:integers\:=}\:\bf{500500}}}}$

★ Solving for first case :-

• AP = 1 + 2 + 3 . . . . . . + n

Therefore,

• n = n
• l = n
• a = 1

★ Using formula :-

$\qquad\qquad\bf\dag\:\bigg({\sf{\green{S_{n} = \dfrac{n}{2}\big\lgroup a + l \big \rgroup}}}\bigg)$

★ Putting all known values :-

$\qquad\leadsto\quad\sf S_{n} = \dfrac{n}{2}\big\lgroup 1 + n \big \rgroup$

$\qquad\leadsto\quad\sf S_{n} = \dfrac{n\big\lgroup 1 + n \big \rgroup}{2}$

$\qquad\leadsto\quad\sf S_{n} = \dfrac{\big\lgroup( 1 \:\times\:n) + (n\:\times\:n) \big \rgroup}{2}$

$\qquad\leadsto\quad{\bf{\red{S_{n} = \dfrac{\big\lgroup n + n^2 \big \rgroup}{2}}}}$

$\small\therefore\:{\underline{\sf{Hence,\:sum\:of\:first\:\bf{n}\:\sf{positive\:integers\:=}\:\bf{\frac{\big\lgroup n + n^2 \big\rgroup}{2}}}}}$

━━━━━━━━━━━━━━━━━━━━━━━━━━