Math, asked by amirtha2828, 9 months ago

can you guys do me a help ... can you add these 1+2+3+4+5+6+7+8+9+0+10+11+12+13+14+15+16+17+18+19+20+381 and put a like to each and every answer and I would also thank yous if you answer it properly I would follow you and like each and every time you answer any questions..​

Answers

Answered by ItzMADARA
3

 \huge \boxed{ \fcolorbox{orange}{pink}{Answer :-}}

Adding 1+2+3+4+5+6+7+8+9+10 up one at a time would take some time, calculating each addition, even if for a few seconds, would take half a minute for some people. So, an easier option has been discovered:Why not make multiple amounts of a certain number, and add them instead? It is a much easier, more fun and faster solution to just adding them all. First, we find the lowest number we can do this with. Some would say 10, as 10 is the lowest number, however that would not work, as we would end with 10 (made of the number 10) X 10 (9 + 1) X 10 (8 + 2 X 10 (7 + 3) X 10 (6 + 4) X … 5? There would still be a five left over, even if the others where all 10’s, which would make 10 X 5 (10 + 10 + 10 + 10 + 10) + 5, easy to do, but wait! Theres yet a faster way! Why not remove the 5 at the end and add it to each of the 10’s, to make an entirely multiple question, quicker and easier to do! This would make 11 X 5 (not using 10’s any more, now 11’s, which still works as 10 + 1 = 11, and 9 + 2 == 11, and so on until 5 + 6 = 11.), without any extra add ons, which is a quick and easy 55. Although it needs multiplication and a little mental work, it proves to be a much faster, and fun, way to solve questions such as that.

P.s: A few of the other questions would-be 1+2+3+4+5…+98+99+100 (adding every number up to 100). The clear answer would-be 101 (100 + 1) + 101 (99 + 2) and so forth until + 101 (50 + 51), which would imply incorporating 50 101’s, or 101 X 50, which any professional multipliers (or calculators) should corresponding to 5050! (near 55, with two 5’s, except it is to 100 perhaps not 10, so its two 50’s instead!). P.p.s: For those of you desiring a great formula with this equation, I have gone and found one: S = n(n+1)\/2, a neat bit of mathematics that you ought to test time!

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