Question

Prove that 1 + 3 + 5 + … + (2n – 1) = n2 using the principle of Mathematical induction.​

Answer 1

Step-by-step explanation:

Statement -

1 + 3 + 5 + … + (2n – 1) = n²

Consider that -

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

[ P(1) is true ∵ P(1) : 1 = 12 ]

Consider P(k) is true for k ∈ N,

This indicates that,

$\sf{P(k) : 1 + 3 + 5+...+ (2k - 1) = {k}^{2}}$

Prove that P(k + 1) is true -

$\sf{\longrightarrow} \: 1 + 3 + 5+...+ (2k - 1) + (2k + 1) \\ \\\sf{\longrightarrow} \: k^{2} + (2k + 1) \\ \\ \sf{\longrightarrow}\:{k}^{2} + 2k + 1 \\ \\\sf{\longrightarrow}\:k(k + 1) + 1(k + 1) \\ \\ \sf{\longrightarrow}\:(k + 1)(k + 1) \\ \\ \sf{\longrightarrow}\:{(k + 1)}^{2}$

Following the formula, the above can be written as -

$\sf{\longrightarrow} \: (k + 1)^{2}$

Whenever P(k) is true, P(k + 1) is true

$\therefore$ It is proved by the principle of Mathematical induction that P(n) is true for all n ∈ N.

Answer 2

$\mid \overline \mathbf \red{solution}$

S = 1 + 3 + 5 + 7 +...+ (2n-5) + (2n-3) + (2n-1)

We can re-arrange and write

S = (2n-1) + (2n-3) + (2n-5) + ...+ 7 + 5 + 3 + 1

Adding both equations

2S = (2n) + (2n) + (2n) + (2n) +....+ (2n) n terms

2S = (2n)(n) = 2 n²

S = n²

1 + 3 + 5 + 7 +...+ (2n-5) + (2n-3) + (2n-1) = n²