# 1 + 2 + 3 + 4 + ⋯

Bạn đang xem: 1 + 2 + 3 + 4 + ⋯  Let \$\$S = 1 + 2 + ldots + (n-1) + n.\$\$ Write it backwards: \$\$S = n + (n-1) + ldots + 2 + 1.\$\$Add the two equations, term by term; each term is \$n+1,\$ so\$\$2S = (n+1) + (n+1) + ldots + (n+1) = n(n+1).\$\$Divide by 2: \$\$S = fracn(n+1)2.\$\$  My favourite proof is the one given here on umakarahonpo.comOverflow. I"m copying the picture here for easy reference, but full credit goes to lớn Mariano Suárez-Alvarez for this answer. Takes a little bit of looking at it to see what"s going on, but it"s nice once you get it. Observe that if there are n rows of yellow discs, then:

there are a total of 1 + 2 + ... + n yellow discs; every yellow disc corresponds to lớn a unique pair of blue discs, và vice versa;there are \$n+1 choose 2 = frac 12 n(n+1)\$ such pairs.
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edited Aquảng cáo 13 "17 at 12:58
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3 revs, 2 users 83%Vaughn Climenhaga
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\$egingroup\$ great proof!!!! \$endgroup\$–anon Aug 12 "10 at 21:22
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What a big sum! This is one of those questions that have sầu dozens of proofs because of their utility & instructional use. I present my two favorite proofs: one because of its simpliđô thị, and one because I came up with it on my own (that is, before seeing others vị it - it"s known).

The first involves the above sầu picture: In short, note that we want to know how many boxes are in the outlined region, as the first column has 1 box, the second 2, và so on (1 + 2 + ... + n). One way khổng lồ count this quickly is khổng lồ take another copy of this section and attach it below, making a \$n*(n+1)\$ box that has exactly twice as many squares as we actually want. But there are \$n*(n+1)\$ little squares in this area, so our sum is half that: \$\$ 1 + 2 + ... + n = dfracn(n+1)2. \$\$

Second proof, same as the first but a little bit harder & a little bit worse:

Let us take for granted the finite geometric sum \$1 + x + x^2 + ... + x^n = dfracx^n+1 - 1x-1\$ (If you are unfamiliar with this, phản hồi and I"ll direct you to a proof). This is a polynomial - so let"s differentiate it. We get \$\$ 1 + 2x + 3x^2 + ... + nx^n-1 = dfrac (n+1)x^n (x-1) - x^n+1 + 1 (x-1)^2 \$\$ Taking the limit as x approaches 1, we get

\$\$ lim_x lớn 1 dfrac (n+1)x^n (x-1) - x^n+1 + 1 (x-1)^2 = dfrac (n+1) < (n+1)x^n - nx^n-1 > - (n+1)x^n 2(x-1) = \$\$\$\$ lim_x o 1 dfrac (n+1)<(n+1)(n)x^n-1 - n(n-1)x^n-2> - (n+1)(n)x^n-1 2\$\$

where we used two applications of l"Hopital above sầu. This limit exists, and plugging in x = 1 we see that we get \$\$ dfrac12 * (n+1)(n) < (n+1) - (n-1) - 1> = dfrac (n)(n+1)2.\$\$