mathematical induction?

S(n) = 1^2 + 2^2 + ... +n^2 = [n(n+1)(2n+1)]/6

How do i prove that this is true for all positive integers n using mathematical induction?

9 Answers

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  • seah
    Lv 7
    1 decade ago
    Best Answer

    Let say

    S(n) = 1^2 + 2^2 + ... +n^2 = [n(n+1)(2n+1)]/6

    is true for n

    for n+1

    LHS

    S(n+1)

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

    = [n(n+1)(2n+1)]/6 + (n+1)^2

    = (2n^3 + 3n^2 + n)/6 + 6(n^2 + 2n + 1)/6

    = (2n^3 + 9n^2 + 13n + 6)/6

    = (n+1)(n+2)(2n+3)/6

    = [(n+1)((n+1)+1)(2(n+1)+1)]/6

    RHS

    true for n +1

    when n=1,

    S(1) = 1^2 = 1 (LHS)

    S(1) = (1)(1+1)(2x1 + 1)/6 = 1 (RHS)

    So ,use induction mathematic

    S(n) = 1^2 + 2^2 + ... +n^2 = [n(n+1)(2n+1)]/6

    is true for all positive integers n

    Source(s): myseah
  • Anonymous
    1 decade ago

    You need two things: a base case and an "inductive leap of faith". Then you prove both.

    In this particular instance, your base case will be S(1), and your "leap of faith" will be that if it is true for S(n), then it is also true for S(n+1).

    S(1) = 1^2 = 1 * 2 * 3 / 6 = 1.

    Sp. true for S(n), prove for S(n + 1).

    Goal: prove that S(n+1) = (n + 1)(n + 2)(2n + 3) / 6

    = (n^2 + 3n + 2)(2n + 3) / 6

    = (2n^3 + 9n^2 + 13n + 6) / 6

    Start with the definition of S(n+1):

    S(n+1) = S(n) + (n+1)^2 = S(n) + n^2 + 2n + 1

    And then substitute the equation for S(n), using the inductive hypothesis:

    S(n+1) = n(n + 1)(2n + 1) / 6 + n^2 + 2n + 1

    = (2n^3 + 3n^2 + n) / 6 + n^2 + 2n + 1

    = ((2n^3 + 3n^2 + n) + (6n^2 + 12n + 6)) / 6

    = (2n^3 + 9n^2 + 13n + 6) / 6

    By the way, you should make sure you understand this proof, and then try to apply it to something else.

  • 4 years ago

    Proof: By Mathematical Induction.

    P(1): 1^2 = 1 = 1 = 6/6 = 1*2*3/6 = 1(1+1)(2*1+1)/6.

    P(n+1): Assume Inductive Hypothesis: 1^2+2^2+3^2+...+n^2 = n(n+1)(2n+1)/6. We need to prove 1^2+2^2+3^2+...+(n+1)^2 = (n+1)(n+2)(2n+3)/6.

    LHS: 1^2+2^2+3^2+...+(n+1)^2 = 1^2+2^2+3^2+...+n^2+(n+1)^2 = (1^2+2^2+3^2+...+n^2)+(n+1)^2 = n(n+1)(2n+1)/6+(n+1)^2 = (n+1)(n(2n+1)/6+(n+1)) = (n+1)((n(2n+1)+6(n+1))/6) = (n+1)((2n^2+n+6n+6)/6) = (n+1)(2n^2+7n+6)/6 = (n+1)(2n^2+4n+3n+6)/6 = (n+1)(2n(n+2)+3(n+2))/6 = (n+1)(n+2)(2n+3)/6.

    Therefore, 1^2+2^2+3^2+...+n^2 = n(n+1)(2n+1)/6 for all n in the set of positive integers. ∎

  • holdm
    Lv 7
    1 decade ago

    basis. S(1). 1^2 = 1 = 1(2)(3)/6. true.

    Assume S(n). Prove S(n+1)

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

    By induction hypothesis, the first group of terms

    = n(n+1)(2n+1)/6

    n(n+1)(2n+1)/6 + n^2+2n+1 = 1/6(2n^3 + 3n^2 + n + 6n^2+12n+6) = 1/6(2n^3 + 9n^2 + 13n +6)=1/6(n+1)(n+2)(2n+3) which is S(n+1).

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  • Como
    Lv 7
    1 decade ago

    Let P(k) be the proposition that:-

    1² + 2² + ------k² = k.(k + 1).(2k + 1) / 6

    Consider P(1):-

    LHS = 1² = 1

    RHS = (1 x 2 x 3) / 6 = 1

    Thus P(1) is true

    Consider P(k + 1):-

    1² + 2² + --k² + (k + 1)² = (k + 1).(k + 2).(2k + 3)/6

    Now 1² +2² + ----k² = k.(k + 1).(2k + 1) / 6

    Add (k + 1)² to both sides:-

    LHS

    1² + 2² + ---k² + (k + 1)²

    RHS

    k.(k + 1).(2k + 1) / 6 + (k + 1)²

    [ k.(k + 1).(2k + 1) + 6.(k + 1)² ] / 6

    [ (k + 1).[(k).(2k + 1) + 6(k + 1)] / 6

    [ (k + 1).(2k² + 7k + 6 ] / 6

    [(k + 1)(k + 2).(2k + 3) / 6

    Thus P(k) true-->P(k + 1) true

    and P(1) true

    Therefore P(n) is true for all + ve integers.

  • 1 decade ago

    1. base case: prove true for n=1

    S(1)=1^2=1

    [1(1+1)(2(1)+1)]/6=[2*3]/6=6/6=1

    2. induction hypothesis, assume true for n=k

    S(k) = 1^2 + 2^2 + ... +k^2 = [k(k+1)(2k+1)]/6

    3. proof: prove true for n=k+1

    S(k+1)=1^2 + 2^2 + ... + k^2 + (k+1)^2 = [k(k+1)(2k+1)]/6 + (k+1)^2 = [k(k+1)(2k+1)+6(k+1)^2]/6

    then you simplify until you get it to equal:

    [(k+1)(k+2)(2(k+1)+1)]/6

    once you do that, you conclude "S(n) = 1^2 + 2^2 + ... +n^2 = [n(n+1)(2n+1)]/6 is true by induction."

  • Anonymous
    1 decade ago

    3 steps.

    1-Show- show that the expression is true for n=1 (pretty much already done)

    2-Assume- assume that the expression is true for n=k (k(k+1)(2k+1)/6=k^2

    3-Prove- break the equation down using basic concepts from Algebra 1. Factoring, like terms, exponent addition/subtraction, etc.

    I think the rest you can figure out.

  • Dr D
    Lv 7
    1 decade ago

    n = 1

    S(1) = 1 = 1*2*3/6 = 1

    verified for n = 1

    Assume it holds for n = k

    ie S(k) = k*(k+1)*(2k+1)/6

    Let's investitage n = k + 1

    S(k+1) = S(k) + (k+1)^2

    = k*(k+1)*(2k+1)/6 + (k+1)^2

    = (k+1)/6 * [ k*(2k+1) + 6*(k+1) ]

    = (k+1)/6 * [ 2k^2 + 7k + 6 ]

    = (k+1)/6 * [ (k+2)*(2k+3) ]

    = (k+1) * (k+1+1) * (2[k+1]+1) / 6

    = S(k+1)

    verified for k+1

    hence S(n) is true for all n

  • Anonymous
    1 decade ago

    Substitute n+1 for n throughout, and simplify, and show that the result differs from the original result by only the additional term (n+1)^2.

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