UPSEE Solved Model Paper 4

Apply districuting rule $(a+b)^3$ = $a^3+3a^2b+3ab^2+b^3$ a = $k^2$ , b = k = $(k^2)^3 + 3(k^2)^2k+3k^2k^2+k^3$ simplify = $k^6+3k^5+3k^4+k^3$ = $\sum\limits_{k=1}^{\infty} \frac{3k^2+3k+1}{k^6+3k^5+3k^4+k^3}$ for an infinite upper boundary , if $a_n$ → 0 , then $\sum\limits_{n=k}^{\infty} (a_{n+1}-a_n)$ = - $a_k$ $\frac{3k^2+3k-1}{k^6+3k^5+3k^4+k^3}$ = - $\left(\frac{1}{(k+1)^3}\right)-\left(-\frac{1}{k^3}\right)$ $a_{n+1}$ = $-\frac{1}{(k+1)^3}$ $a_n$ = $-\frac{1}{k^3}$ $a_k$ = $a_1$ = $-\frac{1}{1^3}$ $\lim\limits_{k\to\infty}\left(-\frac{1}{k^3}\right)$ $\lim\limits_{x\to a}[c.f(x)]$ = c . $\lim\limits_{x\to a}$ f (x) = $-\lim\limits_{k\to\infty}\left(\frac{1}{k^3}\right)$ $\lim\limits_{k\to a}\left[\frac{f(x)}{g(x)}\right]$ = $\frac{ \lim\limits_{x\to a} f(x) }{ \lim\limits_{x\to a} g(x) }$, $\lim\limits_{x\to a}$ g (x) ≠ 0 With the exception of indeterminate form = $-\frac{ \lim\limits_{k\to\infty} 1 }{ \lim\limits_{k\to\infty} k^3 }$ $\lim\limits_{k\to\infty}$ (1) $\lim\limits_{x\to a} c$ = c = 1 $\lim\limits_{k\to\infty}(k^3)$ [Applying infinity property] $\lim\limits_{x\to\infty} (ax^n+\dots+bx+c)$ = ∞, a > 0 , n is odd = 1 , n = 3 = ∞ = - $\frac{1}{\infty}$ = 0 By the telescoping series test : $\sum\limits_{k=1}^{\infty} \frac{3k^2+3k+1}{k^6+3k^5+3k^4+k^3}$ = $-a_k$ = $-\left(-\frac{1}{1^3}\right)$ = 1 Option D is correct answer