Re: How to calculate quickly this sum 192,255,317....1218,1262?

gnh888@xxxxxxxxx nous a récemment amicalement signifié :
I have to calculate hundreds of sums similar to the one below, with
100's + of numbers per series.

How do I calculate this sum?( using multiplication or factorial etc..)

192,255,317,378,438,497,555,612,668,723,777,830,882,
933,983,1032,1080,1127,1173,1218, 1262

This series of number follow this rule
192 -> 255 = 63
255 -> 317 = 62
317 ->378 = 61
378 ->438 = 60
...
..
1173 ->1218 =45
1218 -> 1262 = 44

What is the name of such a series?

Any chance to explain how you calculate the sum?


u_1 = u_0 + a
u_2 = u_1 + a - 1
u_3 = u_2 + a - 2
....
u_p = u_(p-1) + a - (p-1)

I sum this equalities :
u_p = u_0 + ap - (1+2+3+...+(p-1))
u_p = u_0 + ap - p(p-1)/2
u_p = u_0 -p^2/2 + (a+1/2)p

Now, you have to calculate u_0 + u_1 + ... + u_n :

u_0 = u_0 - 0^2/2 + (a+1/2)0
u_1 = u_0 - 1^2/2 + (a+1/2)1
u_2 = u_0 - 2^2/2 + (a+1/2)2
....
u_n = u_0 - n^2/2 + (a+1/2)n

I sum yhis equalities :
S_n = (n+1)u_0 - (0^2 + 1^2 + .. + n^2)/2 + (a+1/2)(0+1+2+...+n)

S_n = (n+1)u_0 - n(n+1)(2n+1)/12 + (a+1/2)n(n+1)/2

S_n = (n+1)u_0 + (3a+1-n)n(n+1)/6

And this is your result.

Example :
S = 192 + 255 + 317 + 378 + 438 + 497

u_0 = 192
a = 63
n = 5

S = 6*192 + (3*63+1-5)*5*6/6
= 1152 +925
= 2077

--
Patrick




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