The value of 1-1-21-1-31-1-4 - - - -1-1-2019 is

The correct option is C 1010 nbsp; nbsp;(1+12)(1+13)(1+14)× ..×(1+12019) =32× nbsp;43× nbsp;54× ...×20202019 =20202=1010 nbsp; ...

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Byju's Answer

Standard XII

Mathematics

nth Term of A.P

The value of ...

Question

The value of $(1 + \frac{1}{2}) (1 + \frac{1}{3}) (1 + \frac{1}{4}) \times . . \times (1 + \frac{1}{2019})$ is

1009

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1010

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2019

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2020

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Solution

The correct option is B $1010$
$(1 + \frac{1}{2}) (1 + \frac{1}{3}) (1 + \frac{1}{4}) \times . . \times (1 + \frac{1}{2019})$
$= \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times . . . \times \frac{2020}{2019} = \frac{2020}{2} = 1010$

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Similar questions

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\frac{(1 + \frac{1}{2}) (1 + \frac{1}{3}) (1 + \frac{1}{4}) . . . . . (1 + \frac{1}{50})}{(1 - \frac{1}{2}) (1 - \frac{1}{3}) (1 - \frac{1}{4}) . . . . . . (1 - \frac{1}{50})}

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The value of

$\frac{3}{4} \div 2 \frac{1}{4}$ of $\frac{2}{3} - \frac{(\frac{1}{2}) - (\frac{1}{3})}{(\frac{1}{2}) + (\frac{1}{3})} \times 3 \frac{1}{3} + \frac{5}{6}$ is

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The value of (1+12)(1+13)(1+14)×..×(1+12019) is

The correct option is B 1010 (1+12)(1+13)(1+14)×..×(1+12019) =32×43×54×...×20202019=20202=1010

The value of $(1 + \frac{1}{2}) (1 + \frac{1}{3}) (1 + \frac{1}{4}) \times . . \times (1 + \frac{1}{2019})$ is

The correct option is B $1010$
$(1 + \frac{1}{2}) (1 + \frac{1}{3}) (1 + \frac{1}{4}) \times . . \times (1 + \frac{1}{2019})$
$= \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times . . . \times \frac{2020}{2019} = \frac{2020}{2} = 1010$