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Using binomial theorem, prove that 8^(n)-7n always leaves remainder 1 when divided by 49. |
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Answer» Solution :For any TWO numbers a and b if we can FIND the numbers q and r such that `a=bq+r`, then we SAY that b divides a with q as QUOTIENT and r as remaindder. Thus, in order to show that `8^(n)-7n` leaves remainer 1 when divided by 49, we prove that `8^(n)-7n=49k+1`, where k is some natural NUMBER. Now, `8^(n)-7` can be written as `(1+7)^(n)-7n` Now, we can write `(1+7)^(n)+{1+^(n)C_(1)7^(1)+.^(n)C_(2)7^(2)+.^(n)C_(3)7^(3)+ . . .+.^(n)C_(n)7^(n)}-7n` `={1+7n+7^(2).^(n)C_(2)+7^(3).^(n)C_(3)+. . . .+7^(n)}-7n` `=1+7^(2)({.^(n)C_(2)+7.^(n)C_(3)+ . . .+7^(n-2)}` `=1+49{.^(n)C_(2)+7.^(n)C_(3)+ . . .+7^(n-2)}` `therefore8^(n)-7n=49k+1,` where `K=.^(n)C_(2)+7.^(n)C_(3)+ . . .+7^(n-2)` this shows that when `8^(n)-7n` is divided by 49 always leaves remainder as 1. |
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