Question

Given that 'n' is a natural number. Prove that the equation below is true using mathematical induction.


`\displaystyle \large{1 + 3 + 5 + ... + (2n - 1) = {n}^(2) }`
Show your work too - thanks!​

Answer 1

To ProvE :-

• 1 + 3 + 5 + ..... + (2n - 1) = n²

Method :-

If P(n) is a statement such that ,

1. P(n) is true for n = 1
2. P(n) is true for n = k + 1 , when it's true for n = k ( k is a natural number ) , then the statement is true for all natural numbers .

`\sf\to \textsf{ Let P(n) : 1 + 3 + 5 + $\dots$ +(2n-1) = n$^(\sf 2)$ }`

Step 1 : Put n = 1 :-

`\sf\longrightarrow LHS = \boxed{\sf 1 } \\`

`\sf\longrightarrow RHS = n^2 = 1^2 = \boxed{\sf 1 }`

Step 2 : Assume that P(n) is true for n = k :-

`\sf\longrightarrow 1 + 3 + 5 + \dots + (2k - 1 ) = k^2`

• Add (2k +1) to both sides .

`\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+1)=k^2+(2k+1)`

• RHS is in the form of ( a + b)² = a²+b²+2ab .

`\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+1)= (k +1)^2`

• Adding and subtracting 1 to LHS .

`\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+1) + 1 -1 = (k +1)^2 \\`

`\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+2) - 1 = (k +1)^2`

• Take out 2 as common .

`\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+\{2(k+1)-1\}= (k +1)^2`

• P(n) is true for n = k + 1 .

Hence by the principal of Mathematical Induction we can say that P(n) is true for all natural numbers 'n' .

**Edits are welcomed**

Answer 2

see below

Explanation:

we want to prove the following using mathematical induction

`\displaystyle 1 + 3 + 5 + ... + (2n - 1) = {n}^(2)`

keep in mind that Mathematical Induction is a special way of proving things. It has only 2 steps:

1. Show it is true for the first one
2. Show that if any one is true then the next one is true

In fact,if you know about Domino effect . it will be easier to understand because That is how Mathematical Induction works! however let our topic back to the question. Showing the step is easy since we just need to prove the first one i.e n=1 . the second step is bit tricky so we'll handle it later,just a bit information the second step is all about assumption. it'll be required later

Step-1:Show it is true for the first one

`2.1 - 1 \stackrel{?}{ = } {1}^(2)`

`1 \stackrel{ \checkmark}{ = } 1`

Step-2:Show that if any one is true then the next one is true

so assuming it true that n=k.we'd obtain

`\rm\displaystyle 1 + 3 + 5 + ... + (2k- 1) = {k}^(2)`

now let n=k+1 therefore we acquire:

`\rm\displaystyle 1 + 3 + 5 + ... + (2k - 1 )+ (2(k + 1)- 1) = {(k + 1)}^(2)`

simplify which yields:

`\rm\displaystyle 1 + 3 + 5 + ... + (2k - 1 )+ 2k + 1 \stackrel{?}{=} {k}^(2) + 2k + 1`

as I mentioned it's all about assumption therefore
`\displaystyle 1 + 3 + 5 + ... + (2k- 1) = {k}^(2)` Thus,

`\rm\displaystyle {k}^(2) + 2k + 1 \stackrel { \checkmark}{ = } {k}^(2) + 2k + 1`

and we are done!

note: the other user is correct but didn't explain the assuming part which can be misleading