The value of mathop {lim }limits_{n to infty | vedclass

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(D) Let $P = e \cdot a^2 \cdot e^3 \cdot a^4 \cdots e^{n-1} \cdot a^n$. We can group the terms with $e$ and $a$ separately: $P = (e^1 \cdot e^3 \cdots

Vedclass · Accepted Answer

$(ae)^{1/4}$

Vedclass · Answer

(D) Let $P = e \cdot a^2 \cdot e^3 \cdot a^4 \cdots e^{n-1} \cdot a^n$. We can group the terms with $e$ and $a$ separately: $P = (e^1 \cdot e^3 \cdots e^{n-1}) \cdot (a^2 \cdot a^4 \cdots a^n)$. The exponents of $e$ are $1, 3, \dots, n-1$,which is an arithmetic progression with $k = n/2$ terms. The sum is $\frac{n/2}{2}(1 + n-1) = n^2/4$. The exponents of $a$ are $2, 4, \dots, n$,which is an arithmetic progression with $k = n/2$ terms. The sum is $\frac{n/2}{2}(2 + n) = \frac{n(n+2)}{4}$. Thus,$P = e^{n^2/4} \cdot a^{n(n+2)/4}$. We need to find $\lim_{n 	o \infty} P^{\frac{1}{n^2+1}} = \lim_{n 	o \infty} (e^{n^2/4} \cdot a^{(n^2+2n)/4})^{\frac{1}{n^2+1}}$. Taking the limit of the exponent: $\lim_{n 	o \infty} \frac{n^2/4}{n^2+1} = 1/4$ and $\lim_{n 	o \infty} \frac{(n^2+2n)/4}{n^2+1} = 1/4$. Therefore,the limit is $e^{1/4} \cdot a^{1/4} = (ae)^{1/4}$.

The value of $\mathop {\lim }\limits_{n \to \infty } {\left( {e \cdot {a^2} \cdot {e^3} \cdot {a^4} \cdots {e^{n - 1}} \cdot {a^n}} \right)^{\frac{1}{{{n^2} + 1}}}}$ is equal to

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