K -lim x -k-1100x-km-xm-101000 is m -101A- 10B- 102C- 103D- 104

The correct option is C (Divide above and below by x^m, then k=x→∞lim ( Σ _k=1^1000 ( 1+k/x )^m/1+10^1000/x^m ) =1+1+1+…… upto 1000 times/1+0 =1000=10^3 ...

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Standard XII

Mathematics

Logarithmic Differentiation

k =lim x →∞∑k...

Question

$k = l i m x \to \infty (\frac{Σ_{k = 1}^{1000} (x + k)^{m}}{x^{m} + 10^{1000}})$ is (m > 101)

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Solution

The correct option is C
$(D i v i d e a b o v e a n d b e l o w b y x^{m}, t h e n k = l i m x \to \infty (\frac{Σ_{k = 1}^{1000} {(1 + \frac{k}{x})}^{m}}{1 + \frac{10^{1000}}{x^{m}}}) = \frac{1 + 1 + 1 + \dots \dots u p t o 1000 t i m e s}{1 + 0} = 1000 = 10^{3}$

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k=limx→∞(Σ1000k=1(x+k)mxm+101000) is (m > 101)

The correct option is C (Divide above and below by xm, thenk=limx→∞(Σ1000k=1(1+kx)m1+101000xm)=1+1+1+……upto 1000 times1+0=1000=103

$k = l i m x \to \infty (\frac{Σ_{k = 1}^{1000} (x + k)^{m}}{x^{m} + 10^{1000}})$ is (m > 101)

The correct option is C
$(D i v i d e a b o v e a n d b e l o w b y x^{m}, t h e n k = l i m x \to \infty (\frac{Σ_{k = 1}^{1000} {(1 + \frac{k}{x})}^{m}}{1 + \frac{10^{1000}}{x^{m}}}) = \frac{1 + 1 + 1 + \dots \dots u p t o 1000 t i m e s}{1 + 0} = 1000 = 10^{3}$