The sum of the series 11- 2-12- 3-13- 4-1100- 101 is equal to

The correct option is B 100101 11× 2+12× 3+⋯ +1100× 101=2 11× 2+3 22× 3+⋯+101 100100× 101=11 12+12 13+⋯ +1100 1101=1 1101=100101 ...

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Byju's Answer

Standard XI

Mathematics

Method of Difference

The sum of th...

Question

The sum of the series $\frac{1}{(1 \times 2)} + \frac{1}{(2 \times 3)} + \frac{1}{(3 \times 4)} + . . . . . . . + \frac{1}{(100 \times 101)}$ is equal to

\frac{200}{101}

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\frac{100}{101}

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\frac{50}{101}

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\frac{25}{101}

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Solution

The correct option is B $\frac{100}{101}$
$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \dots + \frac{1}{100 \times 101} = \frac{2 - 1}{1 \times 2} + \frac{3 - 2}{2 \times 3} + \dots + \frac{101 - 100}{100 \times 101} = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \dots + \frac{1}{100} - \frac{1}{101} = 1 - \frac{1}{101} = \frac{100}{101}$

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Similar questions

$s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = \frac{3 π}{2},$ then the value of $x^{100} + y^{100} + z^{100} - \frac{9}{x^{101} + y^{101} + z^{101}} =$

Q. If

s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = \frac{3 π}{2}

, the value of

x^{100} + y^{100} + z^{100} - \frac{9}{x^{101} + y^{101} + z^{101}}

Q. If

{sin}^{- 1} x + {sin}^{- 1} y + {sin}^{- 1} z = \frac{3 π}{2}

, the value of

x^{100} + y^{100} + z^{100} - \frac{9}{x^{101} + y^{101} + z^{101}}

is-

Q. lf

{sin}^{- 1} x + {sin}^{- 1} y + {sin}^{- 1} z = \frac{3 π}{2}

, then

x^{100} + y^{100} + z^{100} - \frac{9}{x^{101} + y^{101} + z^{101}} =

Q. If

a = {(100)}^{1 / 100}

and

b = {(101)}^{1 / 101}

then

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The sum of the series 1(1×2)+1(2×3)+1(3×4)+.......+1(100×101) is equal to

The correct option is B 10010111×2+12×3+⋯+1100×101=2−11×2+3−22×3+⋯+101−100100×101=11−12+12−13+⋯+1100−1101=1−1101=100101

The sum of the series $\frac{1}{(1 \times 2)} + \frac{1}{(2 \times 3)} + \frac{1}{(3 \times 4)} + . . . . . . . + \frac{1}{(100 \times 101)}$ is equal to