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

Question

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 + &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;. + n3 = n2(n + 1)2 / 4

Suresh Chand · Accepted Answer

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.

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

Leave your Answer:

Hits Counter

Question

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

Leave your Answer:

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