show that the sum of first …
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Sia ? 6 years, 6 months ago
Let S1 be the sum of the first n even natural numbers.
Then, S1 = 2 + 4 + 6 + ....+ 2n
{tex}\Rightarrow{/tex} S1 = {tex}\frac{n}{2}{/tex}[2 {tex}\times{/tex} 2 + (n - 1)2]
{tex}\Rightarrow{/tex} S1 = {tex}\frac{n}{2}{/tex}[4 + 2n - 2] = n(n + 1) ........(i)
Let the S2 be the sum of the first n odd natural numbers.
Then, S2 = {tex}\frac{n}{2}{/tex}[2 {tex}\times{/tex} 1 + (n - 1)2] = n2
From (i),we have, {tex}{/tex} {tex}{/tex}{tex}S_1=n^2(1+\frac{1}{n})=(1+\frac{1}{n})n^2=(1+\frac{1}{n})S_2{/tex}
Therefore, the sum of first n even natural numbers is equal to (1 + {tex}\frac{1}{n}{/tex}) times the sum of the first n odd natural numbers {tex}=(1+\frac{1}{n})S_2{/tex}
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