Question

I dont get the inductive step, but i get the basis step and can someone explain the inductive i see the solution but still dont get when it come to the inductive step and this is "Principle of Math induction"

Prove that 2 − 2 · 7 + 2 · 72 −· · ·+2(−7)n = (1 −(−7)n+1)/4 whenever n is a nonnegative integer.

Prove that 1 · 1! + 2 · 2!+· · ·+n · n! = (n + 1)! − 1 whenever n is a positive integer.

Answer 1

Answer:---------

Prove that 2 − 2 · 7 + 2 · 7² −· · ·+2(−7)ⁿ = (1 −(−7)ⁿ⁺¹)/4 whenever n is a non negative integer.
Solution. In order to prove this for all integers n ≥ 0, we first prove the basis step P(0) and then prove the
inductive step, that P(k) implies P(k + 1).
Basis step: Now in P(0), the left-hand side has just one term, namely 2, and the right-hand side is
(1−(−7)) / 4 = 8 / 4 = 2. Since 2 = 2, we have verified that P(0) is true.

Inductive step: For the inductive step, we assume that P(k) is true and and derive from it the truth of P(k + 1), which is the equation
2 − 2 · 7 + 2 · 7² − · · · + 2(−7)^k + 2(−7)^k+1 = (1 − (−7)^(k+1)+1 ) / 4
To prove an equation like this, it is usually best to start with the more complicated side and manipulate it until we arrive at the other side. In this case we start on the left. Note that all but the last term constitute precisely the left-hand side of P(k), and therefore by the inductive hypothesis, we can replace it by the right-hand side of P(k). The rest is algebra:
2 − 2 · 7 + 2 · 7² − · · · + 2(−7)^k + 2(−7)^k+1 = (1 − (−7)^k+1) / 4 + 2(−7)^k+1
================================> (1 − (−7)^k+1 + 8(−7)^k+1) / 4
================================> (1 + 7(−7)^k+1 ) / 4
================================> (1 − (−7) · (−7)^k+1 ) / 4
================================> (1 − (−7)^(k+1)+1 ) / 4  
================================> Hence Proved.

Prove that 1 · 1! + 2 · 2!+· · ·+n · n! = (n + 1)! − 1 whenever n is a positive integer.
Let P(n) be " 1*1! + 2*2! + ... + n*n! = (n+1)! - 1 ", where n=1, 2, 3, ...

Basis step: 1*1! = (1+1)! - 1 = 1 => P(1) is true.
Inductive step:
Assume P(n) is true, i.e. 1*1! + 2*2! + ... + n*n! = (n+1)! - 1
Then 1*1! + 2*2! + ... + n*n! + (n+1)*(n+1)!
======================> (n+1)! - 1 + (n+1)*(n+1)!
======================> (n+1)! {1 -1 + (n+1) }
======================> (n+1)! { (n+1) }
======================> (n+1)! * (1+ n+1) - 1  
======================> (n+1)! * (n+2) - 1
======================> (n+2)! - 1
The last equation shows that P(n+1) is true. This completes the inductive step and completes the proof.