Missing step

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Suppose that we want to prove that for every integer nn

1+6+11+16+...+(5n4)=n(5n3)21 + 6 + 11 + 16 + ... + (5n-4) = {n(5n-3) \over 2}A student writes the following outline for the proof of induction:

Assume that

1+6+11+16+...+(5k4)=k(5k3)21 + 6 + 11 + 16 + ... + (5k-4) = {k(5k-3) \over 2}

You aim to prove that

1+6+11+16+...+(5k4)+(5(k+1)4)=(k+1)(5(k+1)3)21 + 6 + 11 + 16 + ... +(5k-4) + (5(k+1)-4) = {(k+1)(5(k+1)-3) \over 2}You can do this by substituting our result for n=kn=k

k(5k3)2+(5(k+1)4)=(k+1)(5(k+1)3)2{k(5k-3) \over 2} + (5(k+1)-4) = {(k+1)(5(k+1)-3) \over 2}And then you can use algebra to get both the sides to the same form.

The student missed a step. Write down the name of the missing step.

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