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alexgriva [62]
3 years ago
11

Use induction to show that 12 + 22 + 32 + ... + n2 = n(n+1)(2n+1)/6, for all n > 1.

Mathematics
1 answer:
dlinn [17]3 years ago
4 0

Answer with Step-by-step explanation:

Let P(n)=1^2+2^2+3^2+.....+n^2=\frac{n(n+1)(2n+1)}{6}

Substitute n=2

Then  P(2)=1+2^2=5

P(2)=\frac{2(2+1)(4+1)}{6}=5

Hence, P(n) is true for n=2

Suppose that P(n) is true for n=k >1

P(k)=1^2+2^2+3^2+...+k^2=\frac{k(k+1)(2k+1)}{6}

Now, we shall prove that p(n) is true for n=k+1

P(k+1)=1^2+2^2+3^2+...+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}

LHS

P(k+1)=1^2+2^2+3^2+.....+k^2+(k+1)^2

Substitute the value of P(k)

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

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

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

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

P(k+1)=(k+1)(\frac{2k^2+4k+3k+6}{6})

P(k+1)=\frac{(k+1)(k+2)(2k+3)}{6}

LHS=RHS

Hence, P(n) is true for all n >1.

Hence, proved

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