Let m be the smallest positive integer such that the 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
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(5)
Coeff. x 2
2 C 2 + 3 C 2 + 4 C 2 + .......... + 49 C 2 + 50 C 2 m 2 = (3n + 1) 51 C 3
3 C 3 + 3 C 2 + 4 C 2 + .......... + 49 C 2 + 50 C 2 m 2 = (3n + 1) 51 C 3
n C r + n C r–1 = n+1 C r ⇒ 50 C 3 + 50 C 2 . m 2 = (3n + 1) 51 C 3
50 C 3 + 50 C 2 + (m 2 –1) 50 C 2 = 3n. 
. 50 C 2 + 51 C 3 ⇒ 51 C 3 + (m 2 – 1) 50 C 2 = 51n . 50 C 2 + 51 C 3
m 2 – 1 = 51n ⇒ m 2 = 51n + 1
min value of m 2 for 51n + 1 is integer for n = 5 (51n + 1 , n = 5)
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