Question

Mathematical induction question 1E plz I am stuck?

Answer 1

`\`

`"Please see the proof below."`

Explanation:

`\`

`"Ok, let's do the induction."`

`"Want to prove by induction:"`

`\qquad \qquad \qquad \quad [ (3, 1), (0, 2) ]^n \ = \ [ ( 3^n, 3^n - 2^n ), ( 0, 2^n ) ], \qquad \qquad "for all" \ n \in ZZ^+.`

`\`

`"Proof."`

`"1) Basis Step:" \qquad \qquad \qquad "Verify result for" \ \ n = 1.`

`"LHS of result, at" \ n = 1: \qquad [ (3, 1), (0, 2) ]^1 \ = \ [ (3, 1), (0, 2) ]; \qquad \qquad \qquad\qquad \qquad (1)`

`"RHS of result at" \ n = 1:"`

# \qquad \qquad \qquad \qquad [ ( 3^1, 3^1 - 2^1 ), ( 0, 2^1 ) ] \ = \ [ (3, 3-2), (0, 2) ] \ = \ [ (3, 1), (0, 2) ]. \qquad \qquad \qquad (2) #

`"As the result in (1) and the result in (2) are equal, we have"`
`"now proven the desired result at" \ \ n = 1. \ "The Basis Step has"`
`"been completed."`

`"2) Induction Step:"`
`"Assume the result is true for" \ n, "show that it's true for" \ n + 1.`

`\mbox{Assume:} \qquad \qquad \qquad \qquad [ (3, 1), (0, 2) ]^n \ = \ [ ( 3^n, 3^n - 2^n ), ( 0, 2^n ) ]. \qquad \qquad \qquad \qquad \qquad \quad (3)`

`\mbox{Using this (if we want), now show:}`

`\qquad \qquad \qquad \qquad \qquad [ (3, 1), (0, 2) ]^{ n + 1 } \ = \ [ ( 3^{ n + 1 }, 3^{ n + 1 } - 2^{ n + 1 } ), ( 0, 2^{ n + 1 } ) ]. \qquad \qquad \qquad \quad \ (4)`

`\mbox{Ok. Let's start working to try to show (4):`

`\qquad \qquad \qquad \qquad \qquad \ [ (3, 1), (0, 2) ]^{ n + 1 } \ = \ [ (3, 1), (0, 2) ]^1 [ (3, 1), (0, 2) ]^n`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ [ (3, 1), (0, 2) ] [ (3, 1), (0, 2) ]^n`

`"By our assumption from (3), we can continue this as:"`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ [ (3, 1), (0, 2) ] [ ( 3^n, 3^n - 2^n ), ( 0, 2^n ) ]`

`"Multiplying out, and simplfying:"`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ [ ( 3 \cdot 3^n, 3 (3^n - 2^n ) + 1 \cdot 2^n ), ( 0, 2 \cdot 2^n ) ]`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ [ ( 3 \cdot 3^{ n + 1 }, 3 \cdot 3^{ n } - 3 \cdot 2^n + 2^n ), ( 0, 2 \cdot 2^n ) ]`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ [ ( 3^{ n + 1 }, 3^{ n + 1 } - 2^n ( 3 - 1 ) ), ( 0, 2^{ n + 1 } ) ]`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ [ ( 3^{ n + 1 }, 3^{ n + 1 } - 2^n ( 2 ) ), ( 0, 2^{ n + 1 } ) ]`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ [ ( 3^{ n + 1 }, 3^{ n + 1 } - 2^{ n + 1 } ), ( 0, 2^{ n + 1 } ) ].`

`"Thus, from the last chain of equations, we have now shown:"`

`\qquad \qquad \qquad \qquad [ (3, 1), (0, 2) ]^{ n + 1 } \ = \ [ ( 3^{ n + 1 }, 3^{ n + 1 } - 2^{ n + 1 } ), ( 0, 2^{ n + 1 } ) ].`

`"This result, which we have just shown, is (4) -- what we wanted"`
`"to show in this Step -- the Induction Step."`

`"Thus we have completed the Induction Step."`

`"Now having completed both the Basis Step and the Induction"`
`"Step:"`
`\qquad \qquad \qquad "we have completed the Proof by Induction."`

`"So we have proved the desired result:"`

`\qquad \qquad \qquad \quad [ (3, 1), (0, 2) ]^n \ = \ [ ( 3^n, 3^n - 2^n ), ( 0, 2^n ) ], \qquad \qquad "for all" \ n \in ZZ^+.`

Answer 2

See below.

Explanation:

`1+nx le (1+x)^n`

for `n=1` follows `1+x le 1+x` true

now supposing that `1+nx le (1+x)^n` we get

`1+(n+1) x le (1+x)^(n+1)` or

`1+nx + x le (1+x)^n(1+x)` or

`1+n x + x le (1+x)^n + x(1+x)^n` or as a consequence of `1+nx le (1+x)^n` we get

`x le x(1+x)^n` which is true for all `x > -1`

concluding, by finite induction, the assertion is true.

NOTE:

If `-1 < x < 0` we have also `x le x(1+x)^n` because `0 < (1+x)^n < 1`