lim n ? ? ( 1 2 - 1 ) ( n - 1 ) + ( 2 2 - 2 ) ( n - 2 ) + … + ( ( n - 1 ) 2 - ( n - 1 ) ) · 1 ( 1 3 + 2 3 + … + n 3 ) - ( 1 2 + 2 2 + … + n 2 ) is equal to: [2024]

Question

$\lim_{n \to \infty} \frac{(1^{2} - 1) (n - 1) + (2^{2} - 2) (n - 2) + \dots + ({(n - 1)}^{2} - (n - 1)) \cdot 1}{(1^{3} + 2^{3} + \dots + n^{3}) - (1^{2} + 2^{2} + \dots + n^{2})} is equal to:$ [2024]

Smart Achievers · Accepted Answer

(2)

Let $L = \lim_{n \to \infty} (1^{2} - 1) (n - 1) + (2^{2} - 2) (n - 2) + \dots \frac{. . . + ({(n - 1)}^{2} - (n - 1)) \cdot 1}{(1^{3} + 2^{3} + . . . + n^{3}) - (1^{2} + 2^{2} + . . . + n^{2})}$

$Numerator = \sum_{r = 1}^{n - 1} [(r^{2} - r) (n - r)] = \sum_{r = 1}^{n - 1} (- r^{3} + r^{2} (n + 1) - n r)$

$= - {(\frac{(n - 1) n}{2})}^{2} + \frac{(n + 1) (n - 1) n (2 n - 1)}{6} - \frac{n^{2} (n - 1)}{2}$

So, $L = \lim_{n \to \infty} \frac{\frac{n (n - 1)}{2} [\frac{- n (n - 1)}{2} + \frac{(n + 1) (2 n - 1)}{3} - n]}{{(\frac{n (n + 1)}{2})}^{2} - \frac{n (n + 1) (2 n + 1)}{6}}$

$= \lim_{n \to \infty} \frac{(n - 1) (- 3 n^{2} + 3 n + 2 (2 n^{2} + n - 1) - 6 n)}{(n + 1) (3 n^{2} + 3 n - 4 n - 2)}$

$= \lim_{n \to \infty} \frac{(n - 1) (n^{2} - n - 2)}{(n + 1) (3 n^{2} - n - 2)}$

$= \lim_{n \to \infty} \frac{n^{3} (1 - \frac{1}{n}) (1 - \frac{1}{n} - \frac{2}{n^{2}})}{n^{3} (1 + \frac{1}{n}) (3 - \frac{1}{n} - \frac{2}{n^{2}})} = \frac{1}{3}$

Solution

Q.
$\lim_{n \to \infty} \frac{(1^{2} - 1) (n - 1) + (2^{2} - 2) (n - 2) + \dots + ({(n - 1)}^{2} - (n - 1)) \cdot 1}{(1^{3} + 2^{3} + \dots + n^{3}) - (1^{2} + 2^{2} + \dots + n^{2})} is equal to:$ [2024]

1 $\frac{3}{4}$

2 $\frac{1}{3}$

3 $\frac{1}{2}$

4 $\frac{2}{3}$

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Solution

Q. limn→∞(12-1)(n-1)+(22-2)(n-2)+…+((n-1)2-(n-1))·1(13+23+…+n3)-(12+22+…+n2) is equal to: [2024]

1 34

2 13

3 12