1-x-2-1-2 x-3 x2-4 x3-A- FalseB- True

The correct option is A False (1 + x)^n = 1 + nx + n(n 1)/2!x^2 + n(n 1)(n 2)/3!x^3............ (1 x)^ 2 = 1 + ( 2)( x) + ( 2)( 3)( x)^2/2! + ( 2)( 3)( 4)/3! ...

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Byju's Answer

Standard XII

Mathematics

Combination

1-x-2=1-2 x+3...

Question

${(1 - x)}^{- 2}$ = 1 - 2 $x$ + 3 $x^{2}$ - 4 $x^{3}$ + .........

True

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False

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Solution

The correct option is B
False

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

$(1 - x)^{- 2} = 1 + (- 2) (- x) + \frac{(- 2) (- 3) (- x)^{2}}{2!} + \frac{(- 2) (- 3) (- 4)}{3!} (- x)^{3} . . . .$

= 1 + 2 $x$ + 3 $x^{2}$ + 4 $x^{3}$ .............

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Similar questions

1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . . \infty; | x | < 1

\int (1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . .) d x (| x | < 1)

\int (1 + 2 x + 3 x^{2} + 4 x^{3} + . . .) d x (0 < | x | < 1)

Q. Show that,

(1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . . .)^{\frac{1}{2}} = \frac{1}{(1 - x)}

for

x \neq 1

Q. Solve

\int (1 + 2 x + 3 x^{2} + 4 x^{3} + . . .) d x (0 < | x | < 1)

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(1−x)−2 = 1 - 2x + 3x2 - 4x3 + .........

The correct option is B False (1+x)n = 1 + nx + n(n−1)2!x2 + n(n−1)(n−2)3!x3............ (1−x)−2=1+(−2)(−x)+(−2)(−3)(−x)22!+(−2)(−3)(−4)3!(−x)3.... = 1 + 2x + 3x2 + 4x3.............

${(1 - x)}^{- 2}$ = 1 - 2 $x$ + 3 $x^{2}$ - 4 $x^{3}$ + .........