The sum displaystyle∑ k =199 log √[3]( k / k +1) is equal to (Assume base of logarithm is 10.)

Question

The sum displaystyle∑ k =199 log √[3]( k / k +1) is equal to (Assume base of logarithm is 10.) (A) 2 (B) -2 (C) (-2/3) (D) (2/3)

Tardigrade Admin · Accepted Answer

S =(1/3)[ log (1/2)+ log (2/3)+ log (3/4)+ ldots ldots+ log (99/100)]=(1/3)[ log (1/2) ⋅ (2/3) ⋅ (3/4) ⋅ ldots ldots ldots (99/100)] S =(1/3) log 10 (1/100)=(-2/3).

Q. The sum $k = 1 \sum 99 lo g 3 \frac{k}{k + 1}$ is equal to (Assume base of logarithm is 10.)

2

-2

$\frac{- 2}{3}$

$\frac{2}{3}$

$S = \frac{1}{3} [lo g \frac{1}{2} + lo g \frac{2}{3} + lo g \frac{3}{4} + \dots\dots + lo g \frac{99}{100}] = \frac{1}{3} [lo g \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \dots\dots\dots \frac{99}{100}]$
$S = \frac{1}{3} lo g_{10} \frac{1}{100} = \frac{- 2}{3}$ .

Q. The sum k=1∑99​log3k+1k​​ is equal to (Assume base of logarithm is 10.)

2

-2

3−2​

32​

S=31​[log21​+log32​+log43​+……+log10099​]=31​[log21​⋅32​⋅43​⋅………10099​] S=31​log10​1001​=3−2​.

Q. The sum $k = 1 \sum 99 lo g 3 \frac{k}{k + 1}$ is equal to (Assume base of logarithm is 10.)

$\frac{- 2}{3}$

$\frac{2}{3}$

$S = \frac{1}{3} [lo g \frac{1}{2} + lo g \frac{2}{3} + lo g \frac{3}{4} + \dots\dots + lo g \frac{99}{100}] = \frac{1}{3} [lo g \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \dots\dots\dots \frac{99}{100}]$
$S = \frac{1}{3} lo g_{10} \frac{1}{100} = \frac{- 2}{3}$ .