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Addition and Subtraction of Algebraic Expression

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Question

Show that :
$n^{2} + (n + 1)^{2} + {(n^{2} + n)}^{2} = {(n^{2} + n + 1)}^{2}$

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Solution

LHS $= n^{2} + {(n + 1)}^{2} + {(n^{2} + n)}^{2}$
$= n^{2} + n^{2} + 2 n + 1 + n^{4} + n^{2} + {2 n}^{3}$
$= n^{4} + {2 n}^{3} + {3 n}^{2} + 2 n + 1$
RHS $= {(n^{2} + n + 1)}^{2}$
$= n^{4} + n^{2} + 1 + {2 n}^{3} + {2 n}^{2} + 2 n$
$= n^{4} + {2 n}^{3} + {3 n}^{2} + 2 n + 1 =$ LHS

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a_{n^{2} + 1} = (n^{2} + 1^{2}) - (n^{2} + 1) + 1 = (n^{2} + n + 1) (n^{2} - n + 1) =

n L t \to \infty [\frac{n + 1}{n^{2} + 1^{2}} + \frac{n + 2}{n^{2} + 2^{2}} + \dots + \frac{n + n}{n^{2} + n^{2}}] =

{lim}_{n \to \infty} ⎡ ⎢ ⎢ ⎣ \frac{1}{n} + \frac{1}{\sqrt{n^{2} - 1^{2}}} + \frac{1}{\sqrt{n^{2} - 2^{2}}} + . . . . + \frac{1}{\sqrt{n^{2} - {(n - 1)}^{2}}} ⎤ ⎥ ⎥ ⎦

Q. The value of

lim n \to \infty [\frac{n}{n^{2} + 1^{2}} + \frac{n}{n^{2} + 2^{2}} + \dots + \frac{n}{n^{2} + n^{2}}]

\frac{1}{1 -} \frac{4}{5 -} \frac{9}{13 -} \frac{64}{25 -} \dots \frac{{(n^{2} - 1)}^{2}}{n^{2} + {(n + 1)}^{2}} = \frac{(n + 1) (n + 2) (2 n + 3)}{6}

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