Math, asked by aliwathoyachsadiq, 8 months ago

Write the1st four term of expension of (1+x/2)^ 10 in assending power of x​

Answers

Answered by tiwanasavitojs
2

Step-by-step explanation:

For +ve power ‘n’ First 3 terms of binomial expansion (a+b)^n =

nC0 (a)^n (b)^0 . . . . 1st term

+nC1(a)^(n-1) (b)^1 . . . . .2nd term

+nC2(a)^(n-2) (b)^2 . . . . .3rd term

Hence, (1 + ax)^n =

nC0 (1)^n (ax)^0 . . . . .1st term = 1

+nC1 (1)^(n-1) (ax)^1 . . . . . . .2nd term =12x

+nC2 (1)^(n-2) ( ax)^2 . . . . . . . .3rd term =64x²

Here, nC0 = n! /(n-0)!*0! = 1

nC1 = n! /(n-1)!*1! = n

nC2 = n! /(n-2)!*2! = n(n-1) /2

So, First term = 1

2nd term = n*ax = 12 x

=> na = 12 . . . . . . . . . . . . .eq (1)

3rd term = n(n-1)/2 * a²x² = 64x²

=> (n² - n ) /2 = 64/ a² . . . . . . . . . . . eq(2)

By eq(1) a = 12/n

=> eq (2) => n² - n = 128n² / 144

=> 1 - 1/n = 128/144

=> 1/n = 1 - (128/144)

=> 1/n = 16 /144 = 1/9

=> n = 9

By eq (1) 9a = 12

=> a = 12/9

=> a = 4/3

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