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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 + 5}C_{r–1,} ^{n + 5}C_{r}, ^{n + 5}C_{r+1},
then we have ^{n + 5}C_{r–1} : ^{n + 5}C_{r} :^{ n + 5}C_{r+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 x^{2} in the expansion of (1 + x)^{2 }+ (1 + x)^{3} + ... + (1 + x)^{49} + (1 + mx)^{50} is (3n + 1)^{ 51}C_{3} for some positive integer n. Then the value of n is (JEE Adv. 2016)
Ans. (5)
Sol. (1 + x)^{2 }+ (1 + x)^{3} + .... + (1 + x)^{49} + (1 + mx)^{50}
Coeff. of x^{2} in the above expansion = Coeff. of x^{3} in (1 + x)^{50 }+ Coeff. of x^{2} in (1 + mx) ^{50}
⇒ ^{50}C_{3 }+ ^{50}C_{2} m^{2 }
∴ (3n + 1) ^{51}C_{3} = ^{50}C_{3} + ^{50}C_{2} m^{2}
Least positive integer m for which n is an integer is m = 16 and then n = 5
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