InterviewSolution
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InterviewSolution
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Prove by the principle of mathematical induction that `(n^5)/5+(n^3)/3+(7n)/(15)`is a natural number for all `n in Ndot` |
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Answer» Let P(n): `(n^(5))/(5)+(n^(3))/(3)+(7n)/(15)` is a natural number. When n=1, `(n^(5))/(5)+(n^(3))/(3)+(7n)/(15)=(1^(5))/(5)+(1^(3))/(3)+(7)/(15)=(1)/(5)+(1)/(3)+(7)/(15)=1`, which is a natual number Hence, P(1) is true……(A) Let P(m) be true `Rightarrow (m^(5))/(5)+(m^(3))/(3)+(7m)/(15)` is a natural number .....(i) To prove P(m+1) is true i.e, `(m+1)^(5)/(5)+(m+1)^(3)/(3)+(7(m+1))/(15)` is natural number .....(ii) Expanding (ii), we get `(1)/(5)(m^(5)+5m^4+10m^(3)+10m^(2)+5m+1)+(1)/(3)(m^(3)+m^(2)+3m+1)+(7)/(15)(m+1)` `=(m^(5))/(5)+(m^(3))/(3)+(7m)/(15)+(m^(4)+2m^(3)+3m^(2)+2m)+(1)/(5)+(1)/(3)+(7)/(15)` `=(m^(5))/(5)+(m^(3))/(3)+(7m)/(15)+(m^(4)+2m^(3)+2m+1)` =a natural number `[therefore m^(5)/(5)+(m^(3))/(3)+(7m)/(15)"is a natural number from (i) "]` Hence P(m+1) is true whenever P(m) is true .....(B) From (A) and (B) it follows that P(n) is true for all natural number n. |
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