Adding Consecutive Numbers Math Trick Problem | Algebra #mathhack #maths #math #justicethetutor

[thegreat]

People who passed 10th grade getting excited as they remember APs:

[WillEnjy]

For those who are curious, this leads to a general formula stating that the sum from 1 to n = (n/2) * (n+1)

[AvaL]

Just use this when u see 1+2+3+•••+n
1/2 × n × (n+1)

[KofisKitty]

Easier format is n • (n+1) /2
n= last number of enumaration

50 • 51= 2550
2550 /2= 1275

[hdgenius]

This is known as arithmetic progression the formula is sn = n/2(a+l)

[KaylaRicco_]

If I was in that situation I’d start adding it one by one 💀💀

Edit: This comment was supposed to be taken with a sense of humour so don’t take it seriously like some folks

[AzaabcG]

There is a official formula called ‘Gauss’s method’ which he officially discovered at age of 8.
In which he says’If sequence is being added by constant numbers(arithmetic progression), then’ Add First and Last members of sequence and divide product by two, therefore multiply it by number of how many members are there’
Substantiation:
1+50=51 51/2=25.5 25.5*50=1275

[Jay_]

The actual formula is "Sn= n/2[2a+(n-1)d]"
n is the number i.e 50,
a is 1st term i.e 1,
d is common difference i.e 1.
50/2[2(1)+(50-1)1]
25[2+49]
25x51
1275
This is actually from a algebra topic called "Arathemic Progression"

[downhillgroovy]

This problem is exactly why math was my worst subject in school

[bonzo]

This is called the Gauss Method, it’s pretty simple to understand and useful

[gtmrox]

I think my brain just went through a mental breakdown

[jasonfyk]

for those who want to understand a little more about the forms in chat:
since this type of number is basically a bunch of simple sums, you don't need to worry about sign complexities.
a = first number
b = last number
c = quantities of numbers to be added



example 1:
1+2+3+4
a=1 b=4 *c=4*


example 2:
25+26+27+28+29
a=25 b=29 *c=5*
(25, 26, 27, 28 and 29 are 5 different numbers so c=5)


basically the equation goes like this:
c*((b/2)+(a/2))
equation for example 1:
4*((4/2)+(1/2))
equation for example 2:
5*((29/2)+(25/2))


other ways of writing this equation formula:
c*((b/2)+(a/2))
c*((b*.05)+(a*0.5))
c*((b+a)/2)
c*((b+a)*0.5)
c/2*(b+a)
c*0.5*(b+a)
(c*(b+a))/2
(c*(b+a))*0.5
((c*b)/2)+((c*a)/2)
((c*b))*0.5)+((c*a))*0.5)
((c*b)+(c*a))/2
((c*b)+(c*a))*0.5
and the list continues with several other ways of writing the same equationary formula..

[leoofficial]

"it's not 1000, it can't be that ez"
"915 too little, seems odd"
"Last one not divisible by 5"
*Picks C*

[asparkdeity]

Proof of general formula:

S = 1+2+….+k
S = k+(k-1)+…+1

Add them together;

2S = (k+1) + (k+1) + … + (k+1) [k times]
2S = k(k+1)

S = k(k+1) / 2

Can generalise this for any arithmetic progression of the form Un = a+(n-1)d. This gives:

Sn = n/2 [2a + (n-1)d]. The above is a special case with a = 1, d = 1, n = k

[hitesh]

A more easy way
.
.
Look at the last digit ( in this case 50)

Let 50=x
*Get x/2
50÷2= 25
*multiply x/2 with x
50×25=1250
*add x/2 to the no. obtained
1250+25=1275
Hope it will prove helpful 🙂

[quokka_yt]

I did it in a different way.

1 + 2 + ... + 49 + 50 is an arithmetic progression with the first term a₁ = 1, and the common difference d = 1.

Now we can use the formula that Sₙ = n ⋅ ((a₁+ aₙ) / 2), so

S₅₀ = 50 ⋅ ((1 + 50) / 2) =
= 50 ⋅ (51 / 2) =
= 50 ⋅ 25, 5 =
= 1275

[divinewillikeagwuonu]

finally a math shortcut on YouTube that is actually short and simple

[eden]

Me who studied arithmetic progression in 9th Grade :
"I'm Speed"

Edit: omg how did i got so many likes!!!

[thetruthhurts]

I think I actually had this question in my SAT's - I scrunched my face up, looked at it sideways, and just guesstimated correctly.

[appuchachuvlogs]

This problem can also be solved by the AP technique: Sn=n/2(2a+(n-1)d)
Sn=50/2(2×1+(50-1)1)
Where d is the common difference a2-a1=1
n=last term hr 50
a=first term hr 1
So Sn=25(51)
Sn=1275