A problem by Kartik Sharma
The sequence a n is defined by {a}_{1} = 0, |{a}_{2}|=|{a}_{1} + 1|,... |{a}_{n}|= |{a}_{n-1} + 1| . If
n a 1 + a 2 + a 3 + . . . . . + a n ≥ − b a
where a and b are positive coprime integers. Find the value of a + b .
The answer is 3.
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1 solution
Firstly, The question was flagged because it was not stated the ∣ a 2 ∣ = ∣ a 1 + 1 ∣ , but instead a 2 = ∣ a 1 + 1 ∣ .
It seems to be that you want this condition, but your solution is not clearly written. I do not understand how "square all equalities" led to the next line.
Secondly, because you are dealing with negative numbers, 2 − 1 = − 2 1 . You need to explain whether a + b = − 1 + 2 or 1 + ( − 2 ) . Typically, it's best to state it as − b a to avoid ambiguity.
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Yes, I agree but sir, if we do that, then the answer will be changed and so, it would be better if you make the changes
The solution can be understood now, I think. Thanks for changing the title!
and after this, you can find the answer easily!!
THE QUESTION AND THE SOLUTION ARE BOTH INSPIRED BY A.ENGEL's PROBLEM SOLVING STRATEGIES!!
First of all, square all the following equalities
{a}_{1} = 0, |{a}_{2}|=| {a}_{1} +1|,...|{a}_{n}+1| = |{a}_{n} +1| ,
and add them. Reduction yields {{a}_{n+1}}^2 = 2(a1+···+an)+n ≥ 0 .
This implies a 1 + ⋅ ⋅ ⋅ + a n ≥ − n / 2 .