Question

answer needed urgently......

Answer 1

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Sum of first p even numbers = 2+4+6+8+.....n numbers
= 2x(1+2+3+4+---n numbers) = 2xn(n+1) /2 = n(n+1) --------(1)

Sum of first odd numbers = 1+3+5+7+ .... n numbers
= [1+2+3+4+5+6+7 ....{n+(n-1) or (2n-1) first natural numbers}] 
  - [2+4+6... (n-1) even numbers]
=[(2n-1)(2n) / 2 ]   - [(n-1)(n)(use result (1)] = [2n² - n] - [n² - n] = n² ---(2)

Now {1+ (1/n)}xsum of first p odd numbers 
= {1+(1/n)}xn² = n²+n = n(n+1) sum of first n even numbers

Answer 2

done!!!
refer to the attachment