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# Integer Answer Type Questions: Mathematical Induction and Binomial Theorem | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

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Q.1. The coefficients of three consecutive terms of (1 + x)n+5 are in the ratio 5 : 10 : 14. Then n = (JEE Adv. 2013)

Ans. (6)

Sol.  Let the coefficients of three consecutive term s of (1 + x)n + 5
be n + 5Cr–1,  n + 5Cr,  n + 5Cr+1,
then we have n + 5Cr–1 : n + 5Cr : n + 5Cr+1 = 5 : 10 : 14

or n – 3r + 6 = 0 ...(1)

Also

or 5n – 12r + 18 = 0 (2)
Solving (1) and (2) we get n = 6.

Q. 2. Let m be th e sm allest posi tive i n teger such th at th e coefficient of x2 in the expansion of (1 + x)+ (1 + x)3 + ... + (1 + x)49  + (1 + mx)50 is (3n + 1) 51C3 for some positive integer n. Then the value of n is (JEE Adv. 2016)

Ans. (5)

Sol.  (1 + x)+ (1 + x)3 + .... + (1 + x)49 + (1 + mx)50

Coeff. of x2 in the above expansion = Coeff. of x3 in (1 + x)50 + Coeff. of x2 in (1 + mx) 50
50C+ 50C2 m
∴ (3n + 1) 51C3 = 50C3 + 50C2 m2

Least positive integer m for which n is an integer is m = 16 and then n = 5

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## Maths 35 Years JEE Main & Advanced Past year Papers

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## Maths 35 Years JEE Main & Advanced Past year Papers

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