Prove that for all integers x and y, if x2 + y2 is even then x + y is even. Using induction prove that 13 + 23 + 33 + ………………. + n3 = n2(n + 1)2 / 4

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Solution I:

x2 + y2 is even

We know that,

(x+y)2 = x2 + 2xy + y2

(x+y)2 = x2 + y2 + 2xy

Since, 2xy is always even and we have given (x2 + y2) is even. Addition of both even number results even number. So,

(x+y)2 is even. Eventually, x + y is also even.

Solution II:

Let P(n) = 13 + 23 + 33 + ………………. + n3

Basic Step:

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

So, P(1) is true

Induction HypothesisL

Suppose P(k) is true.

P(k) = 13 + 23 + 33 + ………………. + k3 = \(\frac{k^2 (k + 1)^2 }{4}\)

Inductive Step:

Now, we wish to show P(k+1) is true for this, we add (k+1)3 on both sides then

13 + 23 + 33 + ………………. + k3 + (k+1)3 = \(\frac{k^2 (k + 1)^2 }{4} + (k+1)^3\)

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

Taking (k+1)2 common from both terms,

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

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

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

= \(\frac{(k + 1)^2  (k + 2)^2}{4}\)

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

Therefore, P(k+1) = \(\frac{(k + 1)^2 (k + 1 + 1)^2}{4}\) is true.

Thus, by mathematical induction, P(n) is true for all n.

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