You need to use mathematical induction to prove the formula for every positive integer n, hence, you need to perform the two steps of the method, such that:
Step 1: Basis: Show that the statement P(n) hold for n = 1, such that:
`2 = 1/2*(5*1-1) => 2 =4/2 => 2=2`
Step 2: Inductive step: Show that if P(k) holds, then also P(k + 1) holds:
`P(k): 2 + 7 + .. + (5k-3) = (k(5k-1))/2` holds
`P(k+1): 2 + 7 + .....
You need to use mathematical induction to prove the formula for every positive integer n, hence, you need to perform the two steps of the method, such that:
Step 1: Basis: Show that the statement P(n) hold for n = 1, such that:
`2 = 1/2*(5*1-1) => 2 =4/2 => 2=2`
Step 2: Inductive step: Show that if P(k) holds, then also P(k + 1) holds:
`P(k): 2 + 7 + .. + (5k-3) = (k(5k-1))/2` holds
`P(k+1): 2 + 7 + .. + (5k-3) + (5k+2) = ((k+1)(5k+4))/2`
You need to use induction hypothesis that P(k) holds, hence, you need to re-write the left side, such that:
` (k(5k-1))/2 + (5k+2) = ((k+1)(5k+4))/2`
`5k^2 - k + 10k + 4 = 5k^2 + 4k + 5k + 4`
You need to add the like terms, such that:
`5k^2 + 9k + 4 = 5k^2 + 9k + 4`
Notice that P(k+1) holds.
Hence, since both the basis and the inductive step have been verified, by mathematical induction, the statement `P(n): 2 + 7 + .. + (5n-3)= (n(5n-1))/2` holds for all positive integers n.
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