The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+12 votes
370 views

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 $\sum\limits_{i=1}^{n} i = \frac{n(n+1)}{2}$
asked in Combinatory by Veteran (59.5k points)
retagged by | 370 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. $\sum_{i=0}^{n-1} (2i+1) = n^2$
$\implies 1 + \sum_{i=1}^{n} (2i+1)= n^2 + 2n + 1$
$\implies \sum_{i=1}^{n} 2i + \sum_{i=1}^{n} 1 = n^2 + 2n$
$\implies 2.\sum_{i=1}^{n} i + n = n^2 + 2n$
$\implies 2.\sum_{i=1}^{n} i = n^2 + n$
$\implies \sum_{i=1}^{n} i =\frac{n. (n+1)}{2}$
answered by Veteran (357k points)
selected by
0
Couldn't understand your explanation arjun sir, what is meant by ith row of the pattern ?

Please elaborate a bit more.

@Arjun sir

@Habibkhan
+1

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

0
Nope, still can't get it.
0
1st part is not being clear. more explanation plz?
+5
0
Not able to understand !
@Sachin sir , Can u pls explain in words ?
0

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.

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)

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

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

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

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

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

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

$\sum_{i=1}^{n}(i)=(n^{2}+n)/2$
answered by Active (4.9k points)


Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

39,440 questions
46,623 answers
139,377 comments
57,022 users