Using mathematical Induction- the numbers a n - s are defined by a 0-1- a n -1-3 n 2- n - a n n - 0 Then a n -A- n3-n2-1B- n3-n2C- n3-n2D- n3-n2-1

The correct option is A n^3 n^2+1 Putting n = 0, 1, 2, ………………. n – 1 nbsp; and adding a_n a_0=3∑ (n 1 nbsp; )^2+∑ n ⇒ a_n=n^3 n^2+1 nbsp; ...

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Inductive Step

Using mathema...

Question

Using mathematical Induction, the numbers $a_{n}^{'} s$ are defined by $a_{0} = 1, a_{n + 1} = 3 n^{2} + n + a_{n} (n \geq 0)$ Then $a_{n} =$

$n^{3} + n^{2} + 1$

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$n^{3} - n^{2} + 1$

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$n^{3} - n^{2}$

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$n^{3} + n^{2}$

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Using mathematical Induction, the numbers $a_{n}^{'} s$ are defined by $a_{0} = 1, a_{n + 1} = 3 n^{2} + n + a_{n} (n \geq 0)$ Then $a_{n} =$

p (n) = {1^{3} + 2^{3} + 3^{3} + . . . . + n^{3} = \frac{n^{2} {(n + 1)}^{2}}{4}}

n^{t h}

lim n \to \infty [\frac{1}{n^{3}} + \frac{2^{2}}{n^{3}} + \frac{3^{2}}{n^{3}} + \dots + \frac{n^{2}}{n^{3}}]

l i m x \to \infty (\frac{1}{n^{3} + 1} + \frac{4}{n^{3} + 1} + \frac{9}{n^{3} + 1} + \dots \dots + \frac{n^{2}}{n^{3} + 1})

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