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Mathematics Question on Sum of First n Terms of an AP

lim⁡n→∞(12−1)(n−1)+(22−2)(n−2)+…+((n−1)2−(n−1))(13+23+…+n3)−(12+22+…+n2)\lim_{n \to \infty} \frac{(1^2 - 1)(n-1) + (2^2 - 2)(n-2) + \ldots + ((n-1)^2 - (n-1))}{(1^3 + 2^3 + \ldots + n^3) - (1^2 + 2^2 + \ldots + n^2)}is equal to:

A

23\frac{2}{3}

B

13\frac{1}{3}

C

34\frac{3}{4}

D

12\frac{1}{2}

Answer

13\frac{1}{3}

Explanation

Solution

lim⁡n→∞∑r=1n−1(r2−r)(n−r)\lim_{n \to \infty} \sum_{r=1}^{n-1} (r^2 - r)(n - r)

=lim⁡n→∞(∑r=1nr3−∑r=1nr2)= \lim_{n \to \infty} \left( \sum_{r=1}^n r^3 - \sum_{r=1}^n r^2 \right)

=lim⁡n→∞∑r=1n−1(−r3+r2(n+1)−nr)= \lim_{n \to \infty} \sum_{r=1}^{n-1} \left( -r^3 + r^2(n + 1) - nr \right)

lim⁡n→∞(n(n+1)22−n(n+1)(2n+1)6−n2(n−1)2)\lim_{n \to \infty} \left( \frac{n(n + 1)^2}{2} - \frac{n(n + 1)(2n + 1)}{6} - \frac{n^2(n - 1)}{2} \right)

Simplify further:

lim⁡n→∞((n−1)n2+(n+1)(n−1)n(2n−1)−n2(n−1)6)\lim_{n \to \infty} \left( \frac{(n - 1)n}{2} + \frac{(n + 1)(n - 1)n(2n - 1) - n^2(n - 1)}{6} \right)

lim⁡n→∞[n(n+1)2+n(n+1)2+2n+13]\lim_{n \to \infty} \left[ \frac{n(n + 1)}{2} + \frac{n(n + 1)}{2} + \frac{2n + 1}{3} \right]

lim⁡n→∞n(n−1)2(−n(n−1)+(n+1)(2n−1))\lim_{n \to \infty} \frac{n(n - 1)}{2} \left( -n(n - 1) + (n + 1)(2n - 1) \right)

=lim⁡n→∞n(n+1)(3n2+3n−4n−2)6= \lim_{n \to \infty} \frac{n(n + 1)(3n^2 + 3n - 4n - 2)}{6}

=lim⁡n→∞(n−1)(−3n2+3+2(2n2+n−1)−6)(n+1)(3n2−n−2)= \lim_{n \to \infty} \frac{(n - 1)(-3n^2 + 3 + 2(2n^2 + n - 1) - 6)}{(n + 1)(3n^2 - n - 2)}

=lim⁡n→∞(n−1)(n2+5n−8)(n+1)(3n2−n−2)=13= \lim_{n \to \infty} \frac{(n - 1)(n^2 + 5n - 8)}{(n + 1)(3n^2 - n - 2)} = \frac{1}{3}