Question

Prove that 1 + 3 + 5 + … + (2n – 1) = n^2 using the principle of Mathematical induction.​

Answer 1

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$\Large\fbox{\color{purple}{QUESTION}}$

Prove that 1 + 3 + 5 + … + (2n – 1) = n^2 using the principle of Mathematical induction.

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➡️Given Statement: 1 + 3 + 5 + … + (2n – 1) = n^2

➡️Assume that P(n) : 1 + 3 + 5 +…+ (2n – 1) = n^2 , for n ∈ N

➡️Note that P(1) is true, since

➡️P(1) : 1 = 1^2

➡️Let P(k) is true for some k ∈ N,

➡️It means that,

➡️P(k) : 1 + 3 + 5 + … + (2k – 1) = k^2

➡️To prove that P(k + 1) is true, we have

➡️1 + 3 + 5 + … + (2k – 1) + (2k + 1)

➡️= k^2 + (2k + 1)

➡️= k^2 + 2k + 1

➡️By using the formula, the above form can be written as:

➡️= (k + 1)^2

➡️Hence, P(k + 1) is true, whenever P(k) is true.

➡️Therefore, P(n) is true for all n ∈ N is proved by the principle of Mathematical induction.

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Answer 2

Answer:

SOLUTION:-

➡️Given Statement: 1 + 3 + 5 + … + (2n – 1) = n²

➡️Assume that P(n) : 1 + 3 + 5 +…+ (2n – 1) = n² , for n ∈ N

➡️Note that P(1) is true, since

➡️P(1) : 1 = 1²

➡️Let P(k) is true for some k ∈ N,

➡️It means that,

➡️P(k) : 1 + 3 + 5 + … + (2k – 1) = k²

➡️To prove that P(k + 1) is true, we have

➡️1 + 3 + 5 + … + (2k – 1) + (2k + 1)

➡️= k² + (2k + 1)

➡️= k² + 2k + 1

➡️By using the formula, the above form can be written as:

➡️= (k + 1)²

➡️Hence, P(k + 1) is true, whenever P(k) is true.

➡️Therefore, P(n) is true for all n ∈ N is proved by the principle of Mathematical induction.✔

Step-by-step explanation:

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