Mathematics: 50+51+52++101=can you me? i do not understand

50+51+52++101=can you me? i do not understand

Ответ:

lilly9240

12.03.2022

Mathematics

Suppose

$S=50+51+52+\cdots+100+101$

At the same time, we can write

$S^*=101+100+99+\cdots+51+50$

Note that S=S^* (just reverse the sum). Let's pair the first terms of and S^* , and the second, and the third, and so on:

$S+S^*=(50+101)+(51+100)+(52+99)+\cdots+(100+51)+(101+50)$

Now, each grouped term in the sum on the right side adds to 151. There are 52 grouped terms on that same side (because there are 50 numbers in the range of integers 51-100, plus 50 and 101), which menas

$S+S^*=52\cdot151$

But S=S^* , as we pointed out, so

$2S=52\cdot151\implies S=\dfrac{52\cdot151}2=3926$

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Ответ:

electrofy456

24.07.2021

Mathematics

Step-by-step explanation:

$here \: we \: go : - \\ \\ (0.05)( - 0.25) = - 0.0125 \\ \\ so \: - 0.0125 \: is \: not \: equal \: to \: - 1.25 \\ \\ hence \: answer \: is \: \: \: \: \: \: false$

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50+51+52++101=can you me? i do not understand

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