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Question
For every natural number n, n(n+1) is always
For every natural number n, n(n+1) is always
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Solution
The correct option is B even
The product of two consecutive numbers isalways even.We can prove it by principle of mathematicalinduction.Let P(n) be the statementP(n): n(n+1) is even. For n=1, P(n)=1(1+1)=2, which is even.i.e., P(1) is true.Assume P(n) is true for n=k,i.e., k(k+1)=k2+k is even. ...(1)We have to prove that P(k+1) is true wheneverP(k) is true.Consider (k+1)((k+1)+1) =(k+1)(k+2) =k2+2k+k+2 =k2+k+2(k+1) =2m+2(k+1) (Using (1)] =2(m+k+1)⇒P(k+1) is a multiple of 2,i.e., P(k+1) is even.Thus, P(k+1) is true whenever P(k) is true.Hence, from the principle of mathematical induction,P(n) is true for every natural number.
The product of two consecutive numbers isalways even.We can prove it by principle of mathematicalinduction.Let P(n) be the statementP(n): n(n+1) is even. For n=1, P(n)=1(1+1)=2, which is even.i.e., P(1) is true.Assume P(n) is true for n=k,i.e., k(k+1)=k2+k is even. ...(1)We have to prove that P(k+1) is true wheneverP(k) is true.Consider (k+1)((k+1)+1) =(k+1)(k+2) =k2+2k+k+2 =k2+k+2(k+1) =2m+2(k+1) (Using (1)] =2(m+k+1)⇒P(k+1) is a multiple of 2,i.e., P(k+1) is even.Thus, P(k+1) is true whenever P(k) is true.Hence, from the principle of mathematical induction,P(n) is true for every natural number.
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