If limn→∞ (n+1)k−1nk+1[(nk+1)+(nk+2)+…+(nk+n)]=33. limn→∞ 1nk+1⋅1k+2k+3k+…+nk, then the integral value of k is equal to _______.

LHSlimn→∞ (n+1)k−1nk+1[nk⋅n+1+2+…+n]=limn→∞ (n+1)k−1nk+1⋅n2k+n(n+1)2=limn→∞ (n+1)k−1⋅n2k+1+1n2nk+1⇒limn→∞ 1+1nk+1+1n2⇒k+12RHS⇒limn→∞ 1nk+11k+2k+…+nk=1k+1LHS = RHS⇒k+12=33⋅1k+1⇒(2k+1)(k+1)=66⇒(k−5)(2k+13)=0⇒k=5 or −132

If limn→∞ (n+1)k−1nk+1[(nk+1)+(nk+2)+…+(nk+n)]=33. limn→∞ 1nk+1⋅1k+2k+3k+…+nk, then the integral value of k is equal to _______.

LHSlimn→∞ (n+1)k−1nk+1[nk⋅n+1+2+…+n]=limn→∞ (n+1)k−1nk+1⋅n2k+n(n+1)2=limn→∞ (n+1)k−1⋅n2k+1+1n2nk+1⇒limn→∞ 1+1nk+1+1n2⇒k+12RHS⇒limn→∞ 1nk+11k+2k+…+nk=1k+1LHS = RHS⇒k+12=33⋅1k+1⇒(2k+1)(k+1)=66⇒(k−5)(2k+13)=0⇒k=5 or −132

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If $lim_{n \to \infty} \frac{(n + 1)^{k - 1}}{n^{k + 1}} [(nk + 1) + (nk + 2) + \dots + (nk + n)] = 33$ . $lim_{n \to \infty} \frac{1}{n^{k + 1}} \cdot [1^{k} + 2^{k} + 3^{k} + \dots + n^{k}]$ , then the integral value of k is equal to _______.

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Detailed Solution

LHS

$\begin{array}{l} lim_{n \to \infty} \frac{(n + 1)^{k - 1}}{n^{k + 1}} [nk \cdot n + 1 + 2 + \dots + n] \\ = lim_{n \to \infty} \frac{(n + 1)^{k - 1}}{n^{k + 1}} \cdot [n^{2} k + \frac{n (n + 1)}{2}] \\ = lim_{n \to \infty} \frac{(n + 1)^{k - 1} \cdot n^{2} (k + \frac{(1 + \frac{1}{n})}{2})}{n^{k + 1}}) \\ \Rightarrow lim_{n \to \infty} (1 + \frac{1}{n}) (k + \frac{(1 + \frac{1}{n})}{2}) \end{array}$

$\Rightarrow (k + \frac{1}{2})$

RHS

$\Rightarrow lim_{n \to \infty} \frac{1}{n^{k + 1}} (1^{k} + 2^{k} + \dots + n^{k}) = \frac{1}{k + 1}$

LHS = RHS

$\begin{array}{l} \Rightarrow k + \frac{1}{2} = 33 \cdot \frac{1}{k + 1} \\ \Rightarrow (2 k + 1) (k + 1) = 66 \\ \Rightarrow (k - 5) (2 k + 13) = 0 \\ \Rightarrow k = 5 or - \frac{13}{2} \end{array}$