Let P (n) be the statement that&nbsp;13+23+…+n3=(n(n+1)2)2&nbsp;for the positive integer n.&nbsp;a) What is the statement P (1)?&nbsp;b) Show that P (1) is true, completing the basis step of the proof.&nbsp;c) What is the inductive hypothesis?&nbsp;d) What do you need to prove in the inductive step?&nbsp;e) Complete the inductive step, identifying where you use the inductive hypothesis.&nbsp;f ) Explain why these steps show that this formula is true whenever n is a positive integer.

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Let P (n) be the statement that&amp;nbsp;13+23+…+n3=(n(n+1)2)2&amp;nbsp;for the positive integer n.&amp;nbsp;a) What is the statement P (1)?&amp;nbsp;b) Show that P (1) is true, completing the basis step of the proof.&amp;nbsp;c) What is the inductive hypothesis?&amp;nbsp;d) What do you need to prove in the inductive step?&amp;nbsp;e) Complete the inductive step, identifying where you use the inductive hypothesis.&amp;nbsp;f ) Explain why these steps show that this formula is true whenever n is a positive integer.

James Obrien · Accepted Answer

a) &amp;nbsp;P(1):13=(1−(1+1)2)2b)&amp;nbsp;for&amp;nbsp;&amp;nbsp;η=113=(22)21=12=1QEDc)&amp;nbsp;Assume&amp;nbsp;&amp;nbsp;P(k)&amp;nbsp;&amp;nbsp;for&amp;nbsp;&amp;nbsp;k≥1:13+23+…+k3=(k(k+1)2)2d) &amp;nbsp;P(k+1):e)&amp;nbsp;((k+1)((k+1)+1)2)2=13+23+…+k3+(k+1)3((k+1)((k+1)+1)2)2=(k(k+1)2)+(k+1)3((k+1)(k+2)2)2=(k2+k2)+4(k+1)34(k+1)(k+2)2⋅(k+1)(k+2)2=k2+k2⋅k2+k2+4(k+1)34k4+6k3+13k2+12k+44=k4+2k3+k2+4k3+12k2+45k4+6k3+13k2+12k+44=k4+6k3+13k2+12k+44f) 1. P(b)for constant b2.∀k≥b(P(k)→P(k+1))

Nick Camelot · Accepted Answer

Step 1:a) The statement P(1) is:P(1): 13=(1(1+1)2)2Step 2:b) To show that P(1) is true, we substitute the values and simplify both sides of the equation:13=(1(1+1)2)21=(22)21=12Since the equation is true, P(1) is true.Step 3:c) The inductive hypothesis is:P(k): 13+23+...+k3=(k(k+1)2)2Step 4:d) In the inductive step, we need to prove that if P(k) is true, then P(k+1) is also true. In other words, we assume the equation holds for a particular value of k and then use it to show that it holds for the next value (k+1).Step 5:e) Let&#039;s assume that P(k) is true for some positive integer k. That is:P(k): 13+23+...+k3=(k(k+1)2)2Now we need to prove that P(k+1) is true, which means:P(k+1): 13+23+...+k3+(k+1)3=((k+1)((k+1)+1)2)2To prove this, we start with the left-hand side (LHS) of P(k+1) and use the inductive hypothesis:13+23+...+k3+(k+1)3=(k(k+1)2)2+(k+1)3=k2(k+1)24+(k+1)3=k2(k+1)2+4(k+1)34=(k+1)2(k2+4(k+1))4=(k+1)2(k2+4k+4)4=(k+1)2(k+2)24=((k+1)(k+2)2)2=((k+1)((k+1)+1)2)2=((k+1)(k+2)2)2Therefore, we have shown that P(k+1) holds true based on the assumption that P(k) is true. This completes the inductive step.Step 6:f) The steps above demonstrate that P(1) is true (basis step) and that if P(k) is true, then P(k+1) is also true (inductive step). By following these steps repeatedly, starting from the basis step and applying the inductive step, we can prove that the equation 13+23+...+n3=(n(n+1)2)2 holds true for any positive integer n.

Mr Solver · Accepted Answer

a) P(1): 13=(1((1+1))2)2b) To show that P(1) is true, we substitute n=1 into the equation and simplify:13=(1((1+1))2)21=(1(2)2)21=(22)21=12Thus, P(1) is true.c) The inductive hypothesis is that for some positive integer k, P(k) is true. In other words, we assume that 13+23+...+k3=(k((k+1))2)2 is true.d) In the inductive step, we need to prove that if P(k) is true, then P(k+1) is also true. We need to show that 13+23+...+k3+(k+1)3=((k+1)((k+1)+1)2)2 holds.e) To complete the inductive step, we assume that P(k) is true, which means:13+23+...+k3=(k((k+1))2)2Then, we add (k+1)3 to both sides of the equation:13+23+...+k3+(k+1)3=(k((k+1))2)2+(k+1)3Expanding and simplifying the right side:(k2(k+1)24)+(k+1)3Factoring out (k+1)2:(k+1)2(k24+(k+1))Simplifying (k24+(k+1)):(k24+4(k+1)4)(k2+4k+44)((k+2)24)Substituting back into the equation:13+23+...+k3+(k+1)3=((k+1)((k+1)+1)2)2(k+1)2((k+2)24)=((k+1)(k+2)2)2(k+1)2((k+2)24)=((k+1)(k+2)2)2Thus, using the inductive hypothesis, we have shown that P(k+1) is true.f) By completing the basis step and the inductive step, we have shown that P(n) is true for any positive integer n. This means that the formula 13+23+...+n3=(n((n+1))2)2holds whenever n is a positive integer.

Let P (n) be the statement that 1^3+2^3+...+n^3=\left(n\frac{(n+1)}{2}\right)

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