Solution to Prove by mathematical induction. 1 + 5 + 9 +...+ (4n-3) = (2n+1)(n-1) for n … - Sikademy
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Prove by mathematical induction. 1 + 5 + 9 +...+ (4n-3) = (2n+1)(n-1) for n ≥ 2

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Let P(n) be the proposition that

1 + 5 + 9 +...+ (4n-3) =n(2n-1)

for n\geq1.

Basis Step

P(1) is true, because 1=1(2(1)-1).

Inductive Step

For the inductive hypothesis we assume that P(k) holds for an arbitrary positive integer k. That is, we assume that

1 + 5 + 9 +...+ (4k-3) =k(2k-1)

Under this assumption, it must be shown that P(k + 1) is true, namely, that

1 + 5 + 9 +...+ (4k-3)+(4(k+1)-3)


is also true.

When we add (4(k + 1)-3) to both sides of the equation in P(k), we obtain

1 + 5 + 9 +...+ (4k-3)+(4(k+1)-3)

=k(2k-1)+(4(k + 1)-3)





This last equation shows that P(k + 1) is true under the assumption that P(k) is true. This completes the inductive step.

We have completed the basis step and the inductive step, so by mathematical induction we know that P(n) is true for all positive integers n. That is, we have proven that

1 + 5 + 9 +...+ (4n-3) =n(2n-1)

for n\geq1.

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