Posted: February 13th, 2023

Expressing √3x + 9 as a(1+bx)n:

We can express the equation √3x + 9 in the form of a(1+bx)n by doing the following steps.

First, square both sides of the equation. We get 3x + 9 = (√3x + 9)2 . Simplifying this expression, we get 3x +9 = 3×2+18x+81. Subtracting 81 from both sides gives us 3x -72 = x2 + 18X.

Now, divide both sides by x and take the square root on each side to obtain √3= 1+bX where b = 6/√ x.

Therefore, we can express √3X +9 as a(1+bX)n , where a=9 and n= 2/√ X .

We can express the equation √3x + 9 in the form of a(1+bx)n by doing the following steps.

First, square both sides of the equation. We get 3x + 9 = (√3x + 9)2 . Simplifying this expression, we get 3x +9 = 3×2+18x+81. Subtracting 81 from both sides gives us 3x -72 = x2 + 18X.

Now, divide both sides by x and take the square root on each side to obtain √3= 1+bX where b = 6/√ x.

Therefore, we can express √3X +9 as a(1+bX)n , where a=9 and n= 2/√ X .

Expanding √3X +9 as an ascending series of power of x up to x³ :

The expansion of √3x + 9 can be given as follows:

√3X +9 = (9)( 1+(6/√ X ) X )²

= (9)(1+(12/ X )- (36/(4 X²))

⇒ 9 [ 1+(12/ X )- (36/(4 X²))]

=> 9[1+(12× 1)-(36× 1 / 4)] —> Taking all terms together in the form ax^n => => ax^0 – bxc^2 –> Here A = 9 , B=-36 & C=4 => => => 90 – 36*1*4 ⇒ 54 ⇒ 54*X^0 Term in 0th degree power is 54 → Thus coefficient for term in 0th degree power is 54

Now another way obtaining coefficients without linearity property is using binomial theorem : ⇒ ⇒ ⁴⁰C₀ * (aⁿ)*(Bⁱ)* ((cᵢ)/ i! j!) Where A = 3 , B = 6 & C = ½ So When i and j are equal to zero then 4C₀ *(A⁰)* (B⁰)*((c⁰)/ ₀! ₀!) => 4C₀ * ((c⁰)/ ₀! ₀!) which again gives us coefficient 54 for term containing zero degree powers Which matches with what we have found earlier using linearity property

Higher order terms : ↓ ↓ For higher order terms like Power of ‘¹’ or ‘²’ etc ..we will use binomial theorem Let’s calculate coefficient for term containing one degree power → → 4C¹ * [(Aˣ)* B**¹] * [ C**ˣ / ¹ ! ¹ ] Where A ==> 3 ; B ==> 6 & c ==> ½ So when i ==> ¹ and j== > ¹ We get : 4C¹ * [(Aˣ)* B**¹] * [ C**ˣ / ¹ ! ¹ ] ⇔ 24 × {[(A*) ×B ** ??? ] }× {[{Half} ** ???]} Half raise to first power is One half Hence our final answer becomes 24 × {[One* Six]}× [[Half]] which simplifies too 24 × Six × Half That is Twenty four multiplied by six multiplied by half Which simplifies further too Twelve multiplied By Three And that simplifies further too Nine So coefficient for term having one Degree Power Is nine Similarly if you want higher order terms like two degrees three degrees etc..You can similarly calculate them using Binomial Theorem

Therefore our expanded series would look something like this : 54 + 9* Xi’ −36/4 Xi²

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