The value of limn-r-1n r1-a na-1-a-r a-1-a na-1- where a - is a constant- is

Let nbsp;L=lim_n→∞∑_r=1^n r^1/a (n^a 1/a+r ^a 1/a )n^a+1 On simplifying, we have L=lim_n→∞1n∑_r=1^n ( ( rn )^1/a+ ( rn ) ^a ) =∫_0^1 ( x^1/a+x^a )dx=aa+1+1a+ ...

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Standard XII

Mathematics

Second Fundamental Theorem of Calculus

The value of ...

Question

The value of $lim n \to \infty \frac{n \sum r = 1 r^{\frac{1}{a}} (n^{a - \frac{1}{a}} + r^{a - \frac{1}{a}})}{n^{a + 1}},$ where $a \in N$ is a constant, is

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Solution

Let $L = lim n \to \infty \frac{n \sum r = 1 r^{\frac{1}{a}} (n^{a - \frac{1}{a}} + r^{a - \frac{1}{a}})}{n^{a + 1}}$
On simplifying, we have
$L = lim n \to \infty \frac{1}{n} n \sum r = 1 ({(\frac{r}{n})}^{\frac{1}{a}} + {(\frac{r}{n})}^{a})$
$= 1 \int 0 (x^{\frac{1}{a}} + x^{a}) d x = \frac{a}{a + 1} + \frac{1}{a + 1} = 1$

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Second Fundamental Theorem of Calculus

Standard XII Mathematics

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The value of limn→∞n∑r=1r1a(na−1a+ra−1a)na+1, where a∈N is a constant, is

Let L=limn→∞n∑r=1r1a(na−1a+ra−1a)na+1 On simplifying, we have L=limn→∞1nn∑r=1((rn)1a+(rn)a) =1∫0(x1a+xa)dx=aa+1+1a+1=1

The value of $lim n \to \infty \frac{n \sum r = 1 r^{\frac{1}{a}} (n^{a - \frac{1}{a}} + r^{a - \frac{1}{a}})}{n^{a + 1}},$ where $a \in N$ is a constant, is