InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Prove that `n^5/5 + n^3/3+(7n)/(15)` is a natural number. |
|
Answer» Consider the given statement `P(n):(n^(5))/(5)+(n^(3))/(3)+(7n)/(15)` is a natural number, for all n `inN`. Step I We observe that P(1) is true. `P(1):((1)^5)/(5)+(1)^(3)/(3)+(7(1))/(15)=(3+5+7)/(15)=(15)/(15)=1`, which is a natural number. Hence, P(1) is true. Step II Assume that P(n) is true, for n=k. `P(1):((k)^5)/(5)+(k)^(3)/(3)+(7k)/(15)` is natural number. Step III Now, to prove P(k+1) is true. `(k+1)^(5)/(5)+(k+1)^(3)/(3)+(7(k+1))/(15)` `=(k^(5)+5k^(4)+10k^(2)+5k+1)/(5)+(k^(3)+1+3k(k+1))/(3)+(7k+7)/(15)` `=(k^(5)+5k^(4)+10k^(2)+5k+1)/(5)+(k^(3)+1+3k^(2)+3k)/(3)+(7k+7)/(15)` `=(k^(5))/(5)+(k^(3))/(3)+(7k)/(15)+(5k^(4)+10k^(3)+10k^(2)+5k+1)/(5)+(3k^(2)+3k+1)/(3)+(7k+7)/(15)` `=(k^(5))/(5)+(k^(3))/(3)+(7k)/(15)+K^(4)+2k^(3)+2k^(3)+2k^(2)+k+k^(2)+(1)/(5)+(1)/(3)+(7)/(15)` `=(k^(5))/(5)+(k^(3))/(3)+(7k)/(15)+K^(4)+2k^(3)+3k^(3)+2k+1`, which is a natural number So, P(k+1) is true, whenever P(k) is true. Hence, P(n) is true. |
|