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  1. Using the pattern we can see that $n^2$ is obtained by summing all the odd numbers from $1$ to $2n-1$. For $i^{th}$ row of the pattern if we sum the dots to the right end and then down we get $2i-1$. Then sum up the values of all rows and we get $n^2$.
  2. $\sum\limits_{i=0}^{n-1} (2i+1) = n^2$

$\implies 1 + \sum\limits_{i=1}^{n} (2i+1)= (n+1)^2 (\text{Substituting $n$ for $n-1$)}$
$\implies 1 + \sum\limits_{i=1}^{n} (2i+1)= n^2 + 2n + 1$
$\implies \sum\limits_{i=1}^{n} 2i + \sum\limits_{i=1}^{n} 1 = n^2 + 2n$
$\implies 2.\sum\limits_{i=1}^{n} i + n = n^2 + 2n$
$\implies 2.\sum\limits_{i=1}^{n} i = n^2 + n$
$\implies \sum\limits_{i=1}^{n} i =\frac{n. (n+1)}{2}$

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first part (a) i m taking 4 odd term 1,3,5,7 . sum of 4 odd term=16     which is same to conclude (no of odd terms)^2=sum of consecutive n odd terms

same as induction given here,

$\sum$(for i=0, (2i+1)=1 , for i=1,(2i+1)=3 , for i=2,(2i+1)=5, for i=3,(2i+1)=7)=16 (i.e nothing but (no of terms)^2=(4)^2=16)

 

(b)since in question they have said to use result from above

i m taking above equation

$\sum_{0}^{n-1}(2i+1)= n^{2}$ ,now breaking it into two parts [1 (for i=0 , (2i+1) gives 1 )]+[$\sum_{1}^{n}(2i+1)$] (here if u carefully see then we have increased lower as well as upper limit by 1)=n^2+2n+1(we added 2n+1 in rhs side because upper limit has increased by 1 so nth term will give us 2n+1 which is added in rhs side)

$1+\sum_{i=1}^{n}(2i+1)=n^{2}+2n+1$

$\sum_{i=1}^{n}(2i+1)=n^{2}+2n$

$\sum_{i=1}^{n}(2i)+\sum_{1}^{n}1=n^{2}+2n$

$\sum_{i=1}^{n}(2i)+n=n^{2}+2n$

$\sum_{i=1}^{n}(2i)=n^{2}+n$

$2\sum_{i=1}^{n}(i)=n^{2}+n$

$\sum_{i=1}^{n}(i)=(n^{2}+n)/2$

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