Results 1 to 4 of 4

Math Help - induction of series P1

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    45

    induction of series P1

    Could someone help me with these induction questions? I only need the bit on proving
    n=k+1 using the assumption n=k
    Prove:
    1^3 + 2^3 + 3^3 +....+n^3 = (1+2+3+...+n)^2 for n = 1,2,3...

    1-2+3-4...+n(-1)^(n+1) = 1/4[1 - (2n+1)(-1)^n] for n = 1,2,3...

    5/(1.2.3) + 6/(2.3.4) +7/(3.4.5) +...+(n+4)/n(n+1)(n+2) = [n(3n+7]/[2(n+1)(n+2)]

    Thank you ! =)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by christina View Post
    Could someone help me with these induction questions? I only need the bit on proving
    n=k+1 using the assumption n=k
    Prove:
    1^3 + 2^3 + 3^3 +....+n^3 = (1+2+3+...+n)^2 for n = 1,2,3...

    1-2+3-4...+n(-1)^(n+1) = 1/4[1 - (2n+1)(-1)^n] for n = 1,2,3...

    5/(1.2.3) + 6/(2.3.4) +7/(3.4.5) +...+(n+4)/n(n+1)(n+2) = [n(3n+7]/[2(n+1)(n+2)]

    Thank you ! =)
    Hello Christina,

    For number 1......

    If.... 1^3+2^3+3^3+....+n^3=(1+2+3+....n)^2

    Then.... 1^3+2^3+3^3+...+n^3+(n+1)^3=(1+2+3+...+n+[n+1])^2

    Proof:

    (1+2+3+...+n)^2+(n+1)^3=[1+2+3+...+n+(n+1)]^2 ?

    (1+2+3+...+n)^2+(n+1)^3=[(1+2+3+...+n)+(n+1)][(1+2+3+...+n)+(n+1)] ?

    (1+2+..+n)^2+(n+1)^3=(1+2+..+n)^2+2(n+1)(1+2+..+n)  +(n+1)^2 ?

    (n+1)^3=(n+1)(n+1)^2=(n+1)\left(\sum_{i=1}^ni+(n+1  )\right) ?

    (n+1)^2=2\frac{n(n+1)}{2}+n+1=(n+1)(n+1) ?

    True.

    Now you only need prove it for n=1,
    since the inductive step has proven that...
    if the formula is true for n=1, then it's true for n=2, 3, 4, 5, 6....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by christina View Post
    Prove:

    1-2+3-4...+n(-1)^(n+1) = 1/4[1 - (2n+1)(-1)^n] for n = 1,2,3...
    1-2+3-4+...+n(-1)^{n+1}=\frac{1}{4}\left(1-(2n+1)(-1)^n\right),\ n=1,\ 2,\ 3... ?

    \left(1-2+3-4+...n(-1)^{n+1}+(n+1)(-1)^{n+2}\right)=\frac{1}{4}\left(1-[2(n+1)+1](-1)^{n+1}\right) ?

    \frac{1}{4}\left(1-(2n+1)(-1)^n\right)+(n+1)(-1)^{n+2}

    =\frac{1}{4}\left(1-(2n+1)(-1)^n+4(n+1)(-1)^{n+1}(-1)\right)

    =\frac{1}{4}\left(1+(2n+1)(-1)^{n+1}+4(n+1)(-1)(-1)^{n+1}\right)

    =\frac{1}{4}\left(1+(-1)^{n+1}(2n+1-4n-4)\right)

    =\frac{1}{4}\left(1-(2n+3)(-1)^{n+1}\right)

    =\frac{1}{4}\left(1-[2(n+1)+1](-1)^{n+1}\right)

    True.

    Finally prove for n=1.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by christina View Post
    Prove:

    5/(1.2.3) + 6/(2.3.4) +7/(3.4.5) +...+(n+4)/n(n+1)(n+2) = [n(3n+7]/[2(n+1)(n+2)]

    Thank you ! =)
    \frac{5}{1(2)3}+\frac{6}{2(3)4}+\frac{7}{3(4)5}+..  .+\frac{n+4}{n(n+1)(n+2)}=\frac{n(3n+7)}{2(n+1)(n+  2)} ?

    \frac{5}{1(2)3}+\frac{6}{2(3)4}+...\frac{n+4}{n(n+  1)(n+2)}+\frac{n+5}{(n+1)(n+2)(n+3)}=\frac{(n+1)[3(n+1)+7]}{2(n+2)(n+3)} ?

    \frac{n(3n+7)}{2((n+1)(n+2)}+\frac{n+5}{(n+1)(n+2)  (n+3)}=\frac{(n+1)[3(n+1)+7]}{2(n+2)(n+3)} ?

    Multiplying all terms by (n+1), multiplying the left fraction by \frac{n+3}{n+3} and the centre fraction by \frac{2}{2}

    \frac{n(n+3)(3n+7)}{2(n+2)(n+3)}+\frac{2(n+5)}{2(n  +2)(n+3)}=\frac{(n+1)^2[3(n+1)+7]}{2(n+2)(n+3)} ?

    Now, the numerators need to be checked for equality...

    n(n+3)(3n+7)+2(n+5)=(n+1)^2[3(n+1)+7] ?

    If you multiply these out, you will find that they both are

    3n^3+16n^2+23n+10

    True.

    now prove for n=1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Strong induction vs. structural induction?
    Posted in the Discrete Math Forum
    Replies: 13
    Last Post: April 21st 2011, 01:36 AM
  2. Replies: 3
    Last Post: September 29th 2010, 07:11 AM
  3. Replies: 10
    Last Post: June 29th 2010, 01:10 PM
  4. Induction of series P2
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: February 6th 2010, 06:06 AM
  5. Mathemtical Induction Proof (Stuck on induction)
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: March 8th 2009, 10:33 PM

Search Tags


/mathhelpforum @mathhelpforum