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+13 votes

Use the patterns given to prove that

  1. $\sum\limits_{i=0}^{n-1} (2i+1) = n^2$
    (You are not permitted to employ induction)

  2. Use the result obtained in (a) to prove that $\sum\limits_{i=1}^{n} i = \frac{n(n+1)}{2}$
in Combinatory by Veteran (52.1k points)
edited by | 489 views

2 Answers

+8 votes
Best answer
$a.$ 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 right end and then down we get $2i-1$. Then sum up the values of all rows and we get $n^2$.

$b.$ $\sum_{i=0}^{n-1} (2i+1) = n^2$
$\implies 1 + \sum_{i=1}^{n} (2i+1)= (n+1)^2 (\text{Substituting $n$ for $n-1$)}$
$\implies 1 + \sum_{i=1}^{n} (2i+1)= n^2 + 2n + 1$
$\implies \sum_{i=1}^{n} 2i + \sum_{i=1}^{n} 1 = n^2 + 2n$
$\implies 2.\sum_{i=1}^{n} i + n = n^2 + 2n$
$\implies 2.\sum_{i=1}^{n} i = n^2 + n$
$\implies \sum_{i=1}^{n} i =\frac{n. (n+1)}{2}$
by Veteran (416k points)
edited by
Couldn't understand your explanation arjun sir, what is meant by ith row of the pattern ?

Please elaborate a bit more.

@Arjun sir


Use the properties of summation..

Σ (1) = n

Σ n  = n(n+1)/2

Σ n = n(n+1)(2n+1) / 6

Σ n3 = n2(n+1)2 / 4

Nope, still can't get it.
1st part is not being clear. more explanation plz?
Not able to understand !
@Sachin sir , Can u pls explain in words ?

This diagram is just an edited version of Sachin's in case someone needs it..

Count the no. of small boxes with same color ..the sequence is like 1,3,5,7... upto 2(n-1)+1

i.e. 1,3,5,7,9... 2n-1

Now when i=0 2i+1=1 and when i=(n-1), 2i+1 = 2n-1 

So how many terms are there from i=0 to i=(n-1)? It is n i.e. the length of each side of the square. How many boxes are there in the square then? n*n = n2.



One doubt is coming in ans

How in 2nd line $1+summation=(n+1)^{2}$

should not it be $1+summation=n^{2}+1??$

0 votes
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)







by Active (4.9k points)

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