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The sum of first n natural number
  • a)
    (n/2) (n + 1)
  • b)
    (n/6) (n + 1) (2n + 1)
  • c)
    [(n/2) (n + 1)]²
  • d)
    None of these.
Correct answer is option 'A'. Can you explain this answer?
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The sum of first n natural numbera)(n/2) (n + 1)b)(n/6) (n + 1) (2n + ...
Sum of First n Natural Numbers. We prove the formula 1+ 2+ ... + n = n(n+1) / 2, for n a natural number. There is a simple applet showing the essence of the inductive proof of this result.
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The sum of first n natural numbera)(n/2) (n + 1)b)(n/6) (n + 1) (2n + ...
Solution:

To find the sum of first n natural numbers, we use the formula:
Sum = n(n+1)/2

Let's simplify each option and check which one gives us the same formula as above.

a) (n/2)(n+1)
= (n^2 + n)/2
= n(n+1)/2
This is the formula for the sum of first n natural numbers. Hence, option A is correct.

b) (n/6)(n+1)(2n+1)
= (n^3 + 3n^2 + 2n)/6
This is not the formula for the sum of first n natural numbers.

c) [(n/2)(n+1)]
= (n^2 + n)/2
= n(n+1)/2
This is the formula for the sum of first n natural numbers. However, the expression is enclosed in square brackets, which is unnecessary.

d) None of these
We have already established that option A is correct. Hence, this option is incorrect.

Therefore, the correct answer is option A, which gives us the formula for the sum of first n natural numbers.
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The sum of first n natural numbera)(n/2) (n + 1)b)(n/6) (n + 1) (2n + 1)c)[(n/2) (n + 1)]²d)None of these.Correct answer is option 'A'. Can you explain this answer?
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