Infinity..........

12−22+32−42+52−62.........{ 1 }^{ 2 }-{ 2 }^{ 2 }+{ 3 }^{ 2 }-{ 4 }^{ 2 }+{ 5 }^{ 2 }-{ 6 }^{ 2 }.........
I am unable to solve this problem. Please help

#Series #Infinity #Divergent #Complex #Riddle

Note by Spandan Barhai
6 years, 5 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list
  • bulleted
  • list

1. numbered
2. list
  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

bro which class r u in

prajwal kavad - 6 years, 5 months ago

Log in to reply

class 9. I got this question in a math quiz...

Spandan Barhai - 6 years, 5 months ago

Thanks Everyone!!!

Spandan Barhai - 6 years, 5 months ago

its easy ,just use a2−b2=[a−b][a+b]a^2-b^2=[a-b][a+b] and take -1 common from all terms and then u will get a AP series

prajwal kavad - 6 years, 5 months ago

when you find the differences of the pairs like 1^2-2^2 and 3^2 -4^2, you get 3 - 7 -11 - 15 -19... so on. So, u can solve it using it as an arithmatic progression.

Revanth B S Kashyap - 6 years, 5 months ago

We can write 1^2-2^2+3^2-4^2+5^2-6^2+.... as (1+2)(1-2)+(3+4)(3-4)+(5+6)(5-6)+... = -(1+2+3+4+5+6+....)

=-(n(n+1)/2)=-n(n+1)/2

Vishal S - 6 years, 5 months ago

Read Grandi's Series for questions like these.

The sum you've written isn't defined, since its a divergent series. But, if you assume it has a value S S , you can find its value.

S=12−22+32−42… S = 1^2 - 2^2 + 3^2 - 4^2 \dots

S=02+12−22+32… S = 0^2 + 1^2 - 2^2 + 3^2 \dots

2S=1−3+5−7+… 2S = 1 - 3 + 5 - 7 + \dots .(by adding downwards)

Define T=1−3+5−7+… T = 1 - 3 + 5 - 7 + \dots

Therefore 2S=T 2S = T , now

T=1−3+5−7+… T = 1 - 3 + 5 - 7 + \dots

T=0+1−3+5+… T = 0 + 1 - 3 + 5 + \dots

2T=1−2+2−2+2+… 2T = 1 - 2 + 2 - 2 + 2 + \dots (adding downwards)

2T=1−2(1−1+1−1+… ) 2T = 1 - 2(1 - 1 + 1 - 1 + \dots)

2T=1−2(V) 2T = 1 - 2(V)

Where V V is Grandi's series, which has a value of 12 \frac{1}{2}

∴2T=1−2(12)\therefore 2T = 1 - 2(\frac{1}{2})

  ⟹  2T=0 \implies 2T = 0

  ⟹  T=0 \implies T = 0

Now, 2S=T 2S = T

  ⟹  2S=0 \implies 2S = 0

  ⟹  S=0 \implies S = 0

  ⟹  12−22+32−42⋯=0 \implies1^2 - 2^2 + 3^2 - 4^2 \dots = 0

Siddhartha Srivastava - 6 years, 5 months ago

Log in to reply

Thank you

Spandan Barhai - 6 years, 5 months ago

Why is (1−1+1−1+…)=12( 1 - 1 + 1 - 1 + \ldots ) = \frac {1 }{2}
It is really nonsense

Poetri Sonya - 6 years, 5 months ago

Log in to reply

Read the link I gave in the original post. It deals with the same question. TL;DR of the link,

The series 1 − 1 + 1 − 1 + … has no sum.

...but its sum should be 1/2

Simple way to see this, Let the sum be A A .

A=1−1+1−1+… A = 1 - 1 + 1 - 1 + \dots

A=1−(1−1+1−1+… A = 1 - (1 - 1 + 1 - 1 + \dots

A=1−A A = 1 - A

2A=1 2A = 1

A=12 A = \frac{1}{2}

Siddhartha Srivastava - 6 years, 5 months ago

Log in to reply

@Siddhartha Srivastava – Isn't that infinite sequence is a geometric sequence?

A= 1 + 1(-1)+1(-1)^2+1(-1)^3 .... and the formula of infinite geometric sequence is S = a/(1-r) where a = 1 and r = -1

therefore :

A = 1/(1 - (-1)) = 1/2

Adrian Joshua - 5 years, 7 months ago

Log in to reply

@Adrian Joshua – Kind of...

The formula of the sum of an infinite geometric sequence is only valid when ∣r∣<1 |r| < 1 . So you can not use it here.

But, since the method used here to find the value of A is similar to how the formula is derived, they both will give the same answer.

Siddhartha Srivastava - 5 years, 7 months ago

12−22+32−42+52−62...=(1−2)(1+2)+(3−4)(3+4)+(5−6)(5+6)+...=−1(1+2+3+4+5+6+...)=−∞1^2-2^2+3^2-4^2+5^2-6^2...\\=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+...\\=-1(1+2+3+4+5+6+...)\\=-\infty

Pranjal Jain - 6 years, 5 months ago

Log in to reply

You could also do this

12−22+32−42+52−62+72… 1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 +7^2 \dots

=12+(−22+32)+(−42+52)−(62+72)… = 1^2 +(-2^2 + 3^2) + (- 4^2 + 5^2) - (6^2 +7^2) \dots

=1+(3−2)(3+2)+(5−4)(5+4)+(7−6)(7+6)… = 1 + (3-2)(3+2) + (5-4)(5+4) + (7-6)(7+6) \dots

=1+5+9+13+… = 1 + 5 + 9 + 13 + \dots

=∞ = \infty

Siddhartha Srivastava - 6 years, 5 months ago

Log in to reply

It depends on the parity of number of terms. If ∞ is taken as even then mine is correct otherwise yours.

Pranjal Jain - 6 years, 5 months ago
×

Problem Loading...

Note Loading...

Set Loading...