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