JEE MainMathematicsBinomial Theorem
Let K_ n = _ r=1 ^ n 3^ r-1 4^ n-r . If the remainder when K_ n is divided by 7 is 1 , then the smallest three-digit value of n is
Options
- A100
- B101
- C103
- D109
Correct answer
C. 103
Step-by-step solution
The given series is K_ n = 4^ n-1 + 3 4^ n-2 + 3² 4^ n-3 + + 3^ n-1 . This is a geometric progression with first term a = 4^ n-1 and common ratio 3 4 . Sum K_ n = 4^ n-1 (1 - ( 3 4 )^ n ) 1 - 3 4 = 4^ n-1 - 3^ n 4 1 4 = 4^ n - 3^ n . We are given that 4^ n - 3^ n 1 7 . Let us check the values of 4^ n - 3^ n 7 for small values of n : For n=1 , 4 - 3 = 1 1 7 . For n=2 , 16 - 9 = 7 0 7 . For n=3 , 64 - 27 = 37 2 7 . For n=4 , 256 - 81 = 175 0 7 . For n=5 , 1024 - 243 = 781 4 7 . For n=6 , 4096 - 729 = 3367 0 7 . By Fe