Question

Hey guys
Here is your question
Show that the sum of the first n even natural numbers is equal to
(1+1/n) times the sum of the first n
odd natural numbers.

Answer 1

Sum of first p even numbers = 2+4+6+8+.....p numbers

= 2x(1+2+3+4+---p numbers) = 2xp(p+1) /2 = p(p+1) --------(1)

Sum of first odd numbers = 1+3+5+7+ .... p numbers

= [1+2+3+4+5+6+7 ....{p+(p-1) or (2p-1) first natural numbers}]

  - [2+4+6... (p-1) even numbers]

=[(2p-1)(2p) / 2 ]   - [(p-1)(p)(use result (1)] = [2p² - p] - [p² - p] = p² ---(2)

Now {1+ (1/p)}xsum of first p odd numbers

= {1+(1/p)}xp² = p²+p = p(p+1) sum of first p even numbers

Answer 2

hey mate here is your answer.....
^^
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