In the Expanswen of \( \left(2 x^{2}+r x+1\right)^{6} \), where \( r \) is a number and \( x \) is a vaslable, the coefficient of \( x^{2} \) and \( x^{4} \) are 27 and \( -192 \) respectively fund The possisie value of \( r \)

Question

In the Expansion of \( \left(2 x^{2}+r x+1\right)^{6} \), where \( r \) is a number and \( x \) is a variable, the coefficient of \( x^{2} \) and \( x^{4} \) are 27 and \( -192 \) respectively find The possible value of \( r \)

Answer 1

Given expression:- (2x² + rx + 1)⁶ 

The general term of this expression = 6!/ (p! q! r!) × 1^p × rx^q  × (2x²)^r 

= 6!/ (p! q! r!) × r^q × 2^r × (x^{q + 2r})

               

Case1:-  

 x^q +2r = x² 

q + 2r = 2 

Also, p + q + r = 6, 

The possible values of p, q, r are (4,2,0) and (5,0,1) 

The coefficient of x² = 27(given)

[{6! / (4! 2! 1!)} × r² × 2⁰]  + [{6! / (5! 0! 1!)} × r⁰ × 2¹] = 27

[{6! / (4! 2!)} × r²] + [{6! / (5! 1!)} × 2] = 27

15r² + 12 = 27

15r² = 15 ∴r = ±1 .... (1)  

Case2:- 

x^{q + 2r} = x¹¹

q + 2r = 11 

Also, p + q + r = 6

The only possible value of p, q, r is (0,1,5) 

The coefficient of x¹¹ = -192

{6! / (0! 1! 5!)} × r¹ × 2⁵ = -192

{6! / 5!} × r × 32 = -192

6r = -6

r = -1 .... (2) 

From (1) and (2), r = -1

Comments

Your answer is also correct!

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Thanksssssss