Prove that- if a-1- 1-ax1-a3x1-a5x- -1-ax-1-a2-a4x2-1-a21-a4-a9x3-1-a21-a41-a6-

⇒ Assume ( 1+ax ) ( 1+ a ^ 3 x ) ( 1+ a ^ 5 x ) .... #160; =1+ A _ 2 x+ A _ 2 x ^ 2 +... Change x into a ^ 2 x then ( 1+ a ^ 3 x ) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VIII

Mathematics

Division of a Polynomial by a Polynomial

Prove that, i...

Question

Prove that, if $a < 1, (1 + a x) (1 + a^{3} x) (1 + a^{5} x) . . . . . .$
$= 1 + \frac{a x}{1 - a^{2}} + \frac{a^{4} x^{2}}{(1 - a^{2}) (1 - a^{4})} + \frac{a^{9} x^{3}}{(1 - a^{2}) (1 - a^{4}) (1 - a^{6})} + . . . . . .$

Open in App

Solution

$\Rightarrow$ Assume $(1 + a x) (1 + a^{3} x) (1 + a^{5} x) . . . .$
$= 1 + A_{2} x + A_{2} x^{2} + . . .$
Change x into $a^{2} x$ then
$(1 + a^{3} x) (1 + a^{5} x) (1 + a^{7} x) . . . .$
$= 1 + A_{1} a^{2} x + A_{2} a^{4} x^{2} + . . .$
$(1 + a x) (1 + A_{1} a^{2} x + A_{2} a^{4} x^{2} + . . .) = 1 + A_{1} x + A_{2} x^{2} + . . .$
$a + A_{1} a^{2} = A_{1}; A_{1} = \frac{a}{1 - a^{2}}$
$A_{1} a^{3} + A_{2} a^{4} = A_{2}; A_{2} = \frac{a^{4}}{(1 - a^{4}) (1 - a^{2})}$
Similarly for others,
$∴ (1 + a x) (1 + a^{3} x) (1 + a^{5} x) . . . . = 1 + \frac{a x}{1 - a^{2}} + \frac{a^{4} x^{2}}{(1 - a^{2}) (1 - a^{4})} + . . . . .$

Suggest Corrections

Similar questions

Q. Let the sequence

a_{1}, a_{2},

________

a_{n}

form an A.P. and let

a_{1} = 0

, prove that

\frac{a_{3}}{a_{2}} + \frac{a_{4}}{a_{3}} + \frac{a_{5}}{a_{4}} +

________

+ \frac{a_{n}}{a_{n - 1}} - a_{2} (\frac{1}{a_{2}} + \frac{1}{a_{3}} + . . . + \frac{1}{a_{n - 2}}) = \frac{a_{n - 1}}{a_{2}} + \frac{a_{2}}{a_{n - 1}}

Q. If

1 + a + a^{2} + a^{3} + . . . . . . . . . + a^{n} = (1 + a) (1 + a^{2}) (1 + a^{4})

then

n

is given by

Represent the following mixed infinite decimal periodic fractions as common fractions:
Simplify the following expressions.
$a^{2} (1 - a^{2})^{- 1 / 2} : (\frac{1}{1 + (a (1 - a^{2})^{- 1 / 2})^{2}} \cdot \frac{(1 - a^{2})^{1 / 2} + a^{2} (1 - a^{2})^{- 1 / 2}}{1 - a^{2}}) \cdot$

Q. Let

a = \frac{1 + i}{2}

then prove that

a^{6} + a^{4} + a^{2} + 1 = 0

Q. If

(1 + x + x^{2}) n = a 0 + a^{1} x + a^{2} x + a^{3} x^{3} + . . . . . . . a (2^{n}) x (2^{n})

then prove that

(a^{0} + a^{3} + a^{6} + . . . . . . . . .) = (a^{1} + a^{4} + a^{7}) = (a^{2} + a^{5} + a^{8}) = 3^{(} n - 1),

Join BYJU'S Learning Program

Prove that, if a<1,(1+ax)(1+a3x)(1+a5x)......=1+ax1−a2+a4x2(1−a2)(1−a4)+a9x3(1−a2)(1−a4)(1−a6)+......

⇒ Assume (1+ax)(1+a3x)(1+a5x).... =1+A2x+A2x2+... Change x into a2x then(1+a3x)(1+a5x)(1+a7x)....=1+A1a2x+A2a4x2+...(1+ax)(1+A1a2x+A2a4x2+...)=1+A1x+A2x2+...a+A1a2=A1;A1=a1−a2A1a3+A2a4=A2;A2=a4(1−a4)(1−a2)Similarly for others,∴(1+ax)(1+a3x)(1+a5x)....=1+ax1−a2+a4x2(1−a2)(1−a4)+.....

Prove that, if $a < 1, (1 + a x) (1 + a^{3} x) (1 + a^{5} x) . . . . . .$
$= 1 + \frac{a x}{1 - a^{2}} + \frac{a^{4} x^{2}}{(1 - a^{2}) (1 - a^{4})} + \frac{a^{9} x^{3}}{(1 - a^{2}) (1 - a^{4}) (1 - a^{6})} + . . . . . .$