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Prove using induction that, for all n ∈ ℕ, the sum of the first n square numbers can be found using the formula: $$1^2 + 2^2 + 3^2 + .. - Leaving Cert Mathematics - Question d - 2020

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Prove using induction that, for all n ∈ ℕ, the sum of the first n square numbers can be found using the formula: $$1^2 + 2^2 + 3^2 + ... + n^2 = \frac{n(n+1)(2n+1)}... show full transcript

Worked Solution & Example Answer:Prove using induction that, for all n ∈ ℕ, the sum of the first n square numbers can be found using the formula: $$1^2 + 2^2 + 3^2 + .. - Leaving Cert Mathematics - Question d - 2020

Step 1

Step P(1):

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Answer

To prove the base case, for n = 1:

LHS=12=1LHS = 1^2 = 1

Using the formula:

RHS=1(1+1)(21+1)6=1236=1RHS = \frac{1(1+1)(2 \cdot 1 + 1)}{6} = \frac{1 \cdot 2 \cdot 3}{6} = 1

Thus, for n = 1, LHS = RHS.

Step 2

Step P(k):

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Answer

Assume the formula holds for n = k:

12+22+32+...+k2=k(k+1)(2k+1)61^2 + 2^2 + 3^2 + ... + k^2 = \frac{k(k+1)(2k+1)}{6}

Step 3

Step P(k + 1):

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Answer

Now prove it for n = k + 1:

LHS=12+22+32+...+k2+(k+1)2LHS = 1^2 + 2^2 + 3^2 + ... + k^2 + (k+1)^2

Using the induction hypothesis:

=k(k+1)(2k+1)6+(k+1)2= \frac{k(k+1)(2k+1)}{6} + (k+1)^2

Combine the terms:

=k(k+1)(2k+1)+6(k+1)26= \frac{k(k+1)(2k+1) + 6(k+1)^2}{6}

Factor out (k + 1):

=(k+1)(k(2k+1)+6(k+1))6= \frac{(k+1)(k(2k+1) + 6(k+1))}{6}

Simplifying further:

=(k+1)(2k2+7k+6)6= \frac{(k+1)(2k^2 + 7k + 6)}{6}

Recognizing that (2k^2 + 7k + 6 = (k+2)(2k+3)$$, we can write:

LHS=(k+1)(k+2)(2k+3)6=RHSLHS = \frac{(k+1)(k+2)(2k+3)}{6} = RHS

Step 4

Concluding statement:

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Answer

Therefore, by mathematical induction, the formula is true for all natural numbers n.

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