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2 \times 3
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a_{i-1}
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Comments
The answer is 1.
Note that if sn=r=1∑nr21∀n∈N−{1}s1=1 then {sn}n=1∞ is a strictly increasing sequence that which is always less than 2, that is, n<m⟹sn<smn,m∈Nsn<2∀n∈N
Here are the proofs:
The first statement is true as sm contains more number of positive terms than sn.
To prove the next assertion, note that 0⟹r21⟹sn⟹sn<r(r−1)<r2∀r∈N−{1}<r(r−1)1∀r∈N−{1}<1+r=2∑nr(r−1)1=1+r=2∑nr−11−r1=2−n1<2∀n∈N
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
The answer is 1.
Note that if sn=r=1∑nr21∀n∈N−{1}s1=1 then {sn}n=1∞ is a strictly increasing sequence that which is always less than 2, that is, n<m⟹sn<smn,m∈Nsn<2∀n∈N
Here are the proofs:
The first statement is true as sm contains more number of positive terms than sn.
To prove the next assertion, note that 0⟹r21⟹sn⟹sn<r(r−1)<r2∀r∈N−{1}<r(r−1)1∀r∈N−{1}<1+r=2∑nr(r−1)1=1+r=2∑nr−11−r1=2−n1<2∀n∈N
Now, 1=s1<s2016<2⟹⌊s2016⌋=1
Note:
s2016<n→∞limsn=6π2<1.645
Refer here for the proof of this.
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Nice solution..+1.I guessed it anyway
Try to answer other problems of the set.You are tooooo good in solving problems.
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Will try after jee advanced :)
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How much did you get in jee-main?
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Not so good... 250.
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That is pretty good. well i have a long way to go.I'm still in 9th std going to the 10th.
I think the answer is 2.
This is again for NMTC lvl 2 2016
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no it is of 2015
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Oh sorry I meant the previous year IE 2015....I know because I appeared...
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even I appeared
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How many could you solve??
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same as you...1 and a half
how much problems could you solve??
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1 and a half....definitely not so good...One of my friends solved about 3 and qualified