mathematical induction problem...help?

prove by mathematical induction 10^n+ 3.4^(n+2) + 5 is divisible by 9

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  • 9 years ago
    Best answer

    Let's take p(1); here, n = 1

    LHS : 10^1 + 3 x 4^(1+2) + 5

    = 10 + 3 x 4^3 + 5

    = 15 + 3 x 64

    = 15 + 192

    = 207

    Now, if LHS is divisible by 9, remainder = 0

    207 / 9 : Q = 23 and R = 0

    Therefore, LHS = RHS and p(1) is true...

    Now we assume that p(k) is true; here, k ∈ N

    So, 10^k + 3 x 4^(k+2) + 5 is divisible by 9

    => 10^k + 3 x 4^(k+2) + 5 = 9m ; let m be any natural number...(it can vary)

    => 10^k = 9m - 3 x 4^(k+2) - 5 ; Subtract 3 x 4^(k+2) + 5 from both sides - (1)

    Now, we've to prove that p(k+1) is true ; here k + 1 = n and k ∈ N

    10^n + 3 x 4^(n+2) + 5

    = 10^(k+1) + 3 x 4^(k+1+2) + 5---{As we said that n = k+1}

    = 10^k x 10 + 3 x 4^(k+3) + 5

    = ( 9m - 3 x 4^(k+2) - 5)x10 + 3x4^(k+3) + 5 ---{We put 10^k from (1) here...}

    = 9(10m) - 3x10 x 4^(k+2) - 50 + 3x4^(k+3) + 5

    = 9(10m) - 3x10x4^(k+2) + 3x4^(k+2)x4 - 45

    = 9(10m) - 3 x 4^(k+2)x(10-4 ) - 45 ---{Take 3 x 4^(k+2) common}

    = 9(10m) - 9(5) - 18 x 4^(k+2) ---{as 45 = 9(5) }

    = 9(10m) - 9(5) - 9(3x4^(k+2)) ---{as 18 = 9x3}

    = 9(10m - 5 - 3x4^(k+2)) ---{Take 9 common from all 3 terms...}

    That's it... Let 10m - 5 - 3^4(k+2) be any variable Natural number, p

    =9p => It's divisible by 9...

    Therefore, p(k+1) is true...

    And by the Principle of Mathematical Induction, and for all true values of k, p(k+1) is true, p(n) is true. This means that the formula is proved!

    Hope this helps...

    Source(s): My love for math and Mathematical Induction...
  • 9 years ago

    Let P(n) be the predicate such that

    10^n + 3(4^(n + 2)) + 5 is divisible by 9

    for all n ≥ 0

    The basic step, prove that P(0) is true

    10^0 + 3(4^(0 + 2)) + 5 = 54

    9 | 54

    so P(0) is true

    Now assume P(k) is true

    then

    10^k + 3(4^(k + 2)) + 5 must be divisible by 5

    Then we must prove that P(k + 1) is also true

    P(k + 1)

    = 10^(k + 1) + 3(4^(k + 3)) + 5

    = (10^k)(10^1) + 3(4^(k + 2))(4^1) + 5

    = 10(10^k) + 3(4^(k + 2))(4) + 5

    = 9(10^k) + (10^k) + 12(4^(k + 2)) + 5

    = 9(10^k) + (10^k) + 3(4^(k + 2)) + 9(4^(k + 2)) + 5

    = [ 9(10^k) + 9(4^(k + 2)) ] + (10^k) + 3(4^(k + 2)) + 5

    = 9 [ (10^k) + (4^(k + 2)) ] + [ (10^k) + 3(4^(k + 2)) + 5 ]

    Since 9 [ (10^k) + (4^(k + 2)) ] is divisible by 9 as it has a factor of 9 when factorized, and we assumed P(k) is true or [ (10^k) + 3(4^(k + 2)) + 5 ] is divisible by 9

    Both the terms in P(k + 1) is divisible 9, which makes P(k + 1) divisible by 9, thus P(k + 1) is valid

    P(k) ==> P(k + 1)

    Hence this concludes the proof.

  • 9 years ago

    Mistake: it should be 10^n+3*4^(n+2)+5 not "3.4^(n+2)".

    Let p(n)="10^n+3*4^(n+2)+5 is divisible by 9".

    p(1):10^4+3*4^3+5

    =10197 is divisible by 9 (1+1+9+7=18)=>

    p(1) is true.

    Assume p(k) is true for k>=1, where k is an integer. Then

    p(k+1):10^(k+1)+3*4^(k+2)+5

    =10*10^k+3[(4^(k+1))*(3+1)]+5

    =9*10^k+9*4^(k+1)+[10^k+3*4^(k+1)+5]

    is divisible by 9 =>

    p(k+1) is true too

    So, p(n) is true for any +ve integer n.

  • Anonymous
    9 years ago

    Mistake: it should be 10^n+3*4^(n+2)+5 not "3.4^(n+2)".

    Let p(n)="10^n+3*4^(n+2)+5 is divisible by 9".

    p(1):10^4+3*4^3+5

    =10197 is divisible by 9 (1+1+9+7=18)=>

    p(1) is true.

    Assume p(k) is true for k>=1, where k is an integer. Then

    p(k+1):10^(k+1)+3*4^(k+2)+5

    =10*10^k+3[(4^(k+1))*(3+1)]+5

    =9*10^k+9*4^(k+1)+[10^k+3*4^(k+1)+5]

    is divisible by 9 =>

    p(k+1) is true too

    So, p(n) is true for any +ve integer n.

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  • camara
    Lv 4
    3 years ago

    you like 2 issues: a base case and an "inductive bounce of religion". then you teach the two. in this actual occasion, your base case would be S(a million), and your "bounce of religion" would be that no count number if this is authentic for S(n), then it is likewise authentic for S(n+a million). S(a million) = a million^2 = a million * 2 * 3 / 6 = a million. Sp. authentic for S(n), teach for S(n + a million). purpose: teach that S(n+a million) = (n + a million)(n + 2)(2n + 3) / 6 = (n^2 + 3n + 2)(2n + 3) / 6 = (2n^3 + 9n^2 + 13n + 6) / 6 start up with the definition of S(n+a million): S(n+a million) = S(n) + (n+a million)^2 = S(n) + n^2 + 2n + a million and then exchange the equation for S(n), utilising the inductive hypothesis: S(n+a million) = n(n + a million)(2n + a million) / 6 + n^2 + 2n + a million = (2n^3 + 3n^2 + n) / 6 + n^2 + 2n + a million = ((2n^3 + 3n^2 + n) + (6n^2 + 12n + 6)) / 6 = (2n^3 + 9n^2 + 13n + 6) / 6 by skill of ways, you ought to be certain you already know this data, and then attempt to stick to it to a minimum of something else.

  • Como
    Lv 7
    9 years ago

    This could be tricky.

    It is even trickier if the presentation of the question is suspect.

    More brackets required.

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