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Show that the sum of first p even numbers is equal to ( 1+1/p) times the sum of first p odd numbers

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Sum of first p even numbers = 2+4+6+8+.....p numbers
2*(1+2+3+4+...p: numbers)=2*fracp(p+1)2= p(p+1) --------(1)
Sum of the first odd numbers is 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]
frac(2p-1)(2p)2-[(p-1)(p)] , using, result, 1=[2p^2-p]-[p^2-p]=p^2............(2)

Now (1+frac1p)*sum, of, first, p, odd,, numbers
1+(frac1p)*p^2=p^2+p=p(p+1), sum, of, first, p ,even, numbers

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Deependra Verma

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