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If the coefficients of x7 in [ax2 + \(\frac{1}{bx}\)]11 and x – 7 in [ax2 + \(\frac{1}{bx^2}\)]11 are equal, then show that ab = 1.

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Suppose x7 occurs in (r + 1)th term of expansion [ax2 + \(\frac{1}{bx^2}\) ] 11

Now, Tr + 1 = 11Cr (ax2 ) 11 – r ( \(\frac{1}{bx} \)) 

Tr + 1 = 11Cr a11 – r b– r x 22 – 3r (i)

 This will contain x7 , if 

22 – 3r = 7

 ⇒ 3r = 15 or r = 5. 

Putting r = 5 in (i), we obtain that coefficient of x7 In the expansion of  [ax2 + \(\frac{1}{bx}\) ] 11 is 11C5 a6 b – 5

 Suppose x-7 occurs in (r + 1)th term of the expansion of  [ax - \(\frac{1}{bx^2}\) ] 11 

Now, Tr + 1 = 11Cr (ax)11 – r (− 1 2 ) 

⇒ Tr + 1 = 11Cr a11 – r (– 1)r b – r x11– 3r (ii)

This will contain x-7 , if

 11 – 3r = - 7 

⇒ 3r = 18 or r = 6

Putting r = 6 in (ii), we obtain that coefficient of x – 7 in the expansion of

 [ax - \(\frac{1}{bx^2}\) ] 11 

11C6 a 5 b – 6 (– 1)6

If the coefficient of x7  [ax - \(\frac{1}{bx^2}\) ] 11  is equal to the coefficient of x – 7 ( [ax - \(\frac{1}{bx^2}\) ] 11 then 

11C5 a 6 b – 5 = 11C6 a5 b – 6 (– 1)6 

11C5 ab = 11C6 

⇒ ab = 1 [∵ 11C5 = 11C6]

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