Answered

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prove that 1+3+5+......2n-1=n²

Sagot :

As given by the question

There are given that the series

[tex]1+3+5+\cdots+(2n-1)=n^2[/tex]

Now,

For step 1:

Put n=1

Then LHS =1

And

[tex]\begin{gathered} R\mathrm{}H\mathrm{}S=(n)^2 \\ =(1)^2 \\ =1 \end{gathered}[/tex]

So,

[tex]\therefore L.H.S=R.H.S[/tex]

P(n) is true for n=1.

Now,

Step 2:

Assume that P(n) istrue for n=k

Then,

[tex]1+3+5+\cdots+(2n-1)=k^2[/tex]

Adding 2k+1 on both sides

So, we get:

[tex]1+3+5\ldots+(2k-1)+(2k+1)=k^2+(2k+1)=(k+1)^2[/tex]

P(n) is true for n=k+1

By the principle of mathematical induction P(n) is true for all natural numbers n.

Hence,

[tex]1+3+5+\cdots+(2n-1)=n^2[/tex]

For all n.

Hence proved.

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