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2 Apr 2023•0 j'aime•3 vues

2 Apr 2023•0 j'aime•3 vues

Signaler

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Thank you. Show that for all natural numbers n, n5 - n is divisible by 5. Solution n^5-n base case = 1^5-1 =0 0 divisibleby 5 now let assume that this is true for n = k thus k^5- k is divisible by 5 now we have to prove that this is true for n=k+1 also thus (k+1)^5- k+1 = k^5+5 k^4+10 k^3+10 k^2+5 k+1 - K-1 =( k^5-k ) + 5 k^4+10 k^3+10 k^2+5 k = 5 * something + 5 ( k^4+2k^3+2k^2+k) = 5* something menas even n = k+1 is also divisible by 5 thus proved..

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- 1. Thank you. Show that for all natural numbers n, n5 - n is divisible by 5. Solution n^5-n base case = 1^5-1 =0 0 divisibleby 5 now let assume that this is true for n = k thus k^5- k is divisible by 5 now we have to prove that this is true for n=k+1 also thus (k+1)^5- k+1 = k^5+5 k^4+10 k^3+10 k^2+5 k+1 - K-1 =( k^5-k ) + 5 k^4+10 k^3+10 k^2+5 k = 5 * something + 5 ( k^4+2k^3+2k^2+k) = 5* something menas even n = k+1 is also divisible by 5 thus proved.