Question

by how much is the sum of the first 50 integers divisible by 9 greater than the sum of the first 50 positive integer divisible by 8


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Answer 1

Answer:

$Sn=(n/2)(a1+an) an=a1+(n−1)d n=50 a1=2 d=2 ⟹ a50=2+(49)2=100 S50=(50/2)(2+100) S50=(25)(102) S50=2550 it’s the same trick Gauss used to count the numbers from 1 to 100 by folding the serie and add up each pair and multiply with half of the length. In this case we get withn=100 a1=1 d=1 ⟹ a100=1+(99)1=100 S100=(100/2)(1+100) S100=(50)(101) S100=5050 or the sum of the first 50 positive odd numbersn=50 a1=1 d=2 ⟹ a50=1+(49)2=99$

Step-by-step explanation: