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prove that (101)^50>(100)^50+(99)650
Hi Raphael,
Well, you know that (101)^50 = (100 + 1)^50 = 100^50 + 50*100^49 +... + 1 by the binomial formula.
So clearly,
50* 100^49 > 99 * 650.
This implies that since
101^50 > 100^50 + 50*100^49,
by transitive property, 101^50 >(100)^50 + 99(650) thus proved.
I hope this helps,
Robi
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Answers by Expert:
Robi Bhattacharjee
Expertise
I can answer a variety of questions on mathematics. Questions on trigonometry, calculus(preferably single variable), algebra, geometry, and number theory will be answered. I cannot answer questions on abstract branches of mathematics such as group theory. I also cannot answer questions on statistics. In number theory, I can answer questions on congruences, prime numbers, units, functions, and the riemann-zeta function.
Experience
I have studied advanced math my entire life. I started calculus in sixth grade. I have attended numerous math competitions and I am attending math organizations such as the San-Diego math circle. Also, this year I have been invited to the USAMO which is a prestigious math competition (Every year the USAMO invites 500 students from across the USA to participate in this competition. The top 6 go to represent the USA in the International Math Olympiad).
Organizations
I am in the San Diego Math Circle
Education/Credentials
I am entering high school and have received a perfect score and the STAR test 5 times in a row. I also have gotten recognitions in the AMC 10, AIME, Math Counts, and ARML. Additionally, I have won the San Diego Math Olimpiad twice in a row.
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