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Thread: Mathematical induction

  1. #1
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    Question Mathematical induction -P1

    Hi,

    Can someone pls help to solve the below question asap.

    Prove the statement is true using the mathematical induction.

    2 + 4 + 6 +..... + 2n = n (n +1)


    Best Regards
    Lalit Chugh[/B]
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  2. #2
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    Quote Originally Posted by lalitchugh
    Hi, Can someone pls help to solve the below question asap.
    Prove the statement is true using the mathematical induction.
    2 + 4 + 6 +..... + 2n = n (n +1)
    Best Regards
    Lalit Chugh[/B]
    Hello,

    I presume, that you know how to do a proof by induction.

    A(1): n = 1. That means your equation becomes: 2 = 1*(1+1) which is true.

    A(2): Let 2+4+6+...+2n = n(n+1) be true.

    A(3): Proof that 2+4+6+...+2n+2(n+1) =n(n+1)+2(n+1). Factor the RHS

    2+4+6+...+2n+2(n+1) =(n+1)(n+2)

    2+4+6+...+2n+2(n+1) =(n+1)(n+1+1)

    That means you've changed successfully the n's at the RHS into (n+1)'s. So this equation is true for all n\in \mathbb{N}

    Greetings

    EB
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  3. #3
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    Hello, Lalit!

    Are you a bit shaky on Induction?
    This one is fairly simple . . .


    Prove the statement is true using the mathematical induction.
    . . . 2 + 4 + 6 + \hdots + 2n \;= \;n(n +1)

    Verify S(1):\;\;2 \:= \:1(1 + 1) \:= \:2 . . . True!


    Assume S(k):\;\; 2 + 4 + 6 + \hdots + 2k \;= \; k(k + 1)


    Add 2(k + 1) to both sides:

    . . . 2 + 4 + 6 + \hdots + 2k + 2(k + 1) \;= \;k(k + 1) + 2(k+1)

    . . . 2 + 4 + 6 + \hdots + 2(k+1) \;= \;k^2 + k = 2k + 2

    . . . 2 + 4 + 6 + \hdots + 2(k+1) \;=\;(k+1)(k+2) . . . This is S(k+1)\,!

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