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  • 10. If the coefficient of second, third and fourth terms in the expansion of 1 plus x exponent of 2 are in A.P., then show that 2n exponent of 2 minus 9n plus 7 is equal to 0

If the coefficient of second, third and fourth terms in the expansion of (1 + x)2” are in A.P., then show that 2n2 – 9n + 7 = 0.

 

Answers (1)

Given: (1+x)2n

Now, 2nC1, 2nC2, 2nC3 are the coefficients of the first second and the third terms respectively.

It is also given that the coefficients are in A.P.

Thus,  2.  2nC2 =2nC1 + 2nC3

Thus, 2[2n(2n-1)(2n-2)!/2x1x(2n-2)!] = 2n + 1n(2n-1)(2n-2)(2n-3)!/3!(2n-3)!

N(2n-1) = n = n(2n-1)(n-1)/3

3(2n-2) = 3+ (2n2 – 3n + 1)

6n – 3 = 2n2 – 3n + 4

Thus, 2n2 – 9n + 7 = 0

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