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Using binomial theorem show that 1^(99) + 2^(99) +3^(99) + 4^(99) + 5^(99) is divisible by 5 |
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Answer» Solution :`1^(99) + 2^(99) + 3^(99) + 4^(99) + 5^(99)` = 1+ (5-3)^(99) + 3^(99) + (5-1)^(99) + 5^(99) = 1 + (5^99- "^(99)C_1 5^(98). 3^1 + ^(99)C_2 5^(97). 3^2 - ... 3^(99)) + 3^99 - (1- ^(99)C_1 5^1 + ^(99)C_2 5^2 - ... - 5^(99)) + 5^(99)` = `(3xx5^(99) - "^(99)C_1 5^(98). 3^1 + ^(99)C_2 5^(97). 3^2 - .... + ^(99)C_(98) 5^1. 3^(98)) + (^(99)C_15^1 - ^(99)C_2 5^2 + ... -^(99)C_98 5^(98)) ... (1) which is divisible by 5 as each term is a multiple of 5 |
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