1-1-3x-1-4-3-6x2-1-4-7-3-6-9x3- is equal to

The correct option is D (1 x) ^ 13 The above expression can be written as =1+13(x)+13(13+1)2!x^2+13(13+1)(13+2)3!x^3+...∞ =1(1 x)^13 =(1 x) ...

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

1+1/3x+1×4/3×...

Question

$1 + \frac{1}{3} x + \frac{1 \times 4}{3 \times 6} x^{2} + \frac{1 \times 4 \times 7}{3 \times 6 \times 9} x^{3} + . . .$ is equal to

x

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(1 + x)^{\frac{1}{3}}

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(1 - x)^{\frac{1}{3}}

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(1 - x)^{- \frac{1}{3}}

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Solution

The correct option is D $(1 - x)^{- \frac{1}{3}}$
The above expression can be written as

$= 1 + \frac{1}{3} (x) + \frac{\frac{1}{3} (\frac{1}{3} + 1)}{2!} x^{2} + \frac{\frac{1}{3} (\frac{1}{3} + 1) (\frac{1}{3} + 2)}{3!} x^{3} + . . . \infty$
$= \frac{1}{(1 - x)^{\frac{1}{3}}}$
$= (1 - x)^{\frac{- 1}{3}}$

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Similar questions

lim x \to 1 \frac{1 - x^{\frac{1}{3}}}{1 - x^{\frac{1}{4}}} =

f (x) = 9 x^{3} + 6 x^{2} + 3 x + 1

find the value if

x = 3

Q. In the expansion of

{⎛ ⎜ ⎝ \frac{x + 1}{x^{\frac{2}{3}} - x^{\frac{1}{3}} + 1} - \frac{x - 1}{x - x^{\frac{1}{2}}} ⎞ ⎟ ⎠}^{10}

, the term which does not contain

x

, is equal to:

Solve :

$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$

Q. When

x^{13} + 1

is divided by

x - 1

, the remainder is

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1+13x+1×43×6x2+1×4×73×6×9x3+... is equal to

The correct option is D (1−x)−13The above expression can be written as=1+13(x)+13(13+1)2!x2+13(13+1)(13+2)3!x3+...∞=1(1−x)13=(1−x)−13

$1 + \frac{1}{3} x + \frac{1 \times 4}{3 \times 6} x^{2} + \frac{1 \times 4 \times 7}{3 \times 6 \times 9} x^{3} + . . .$ is equal to