Function

Question

The function defined for positive integers by $F(1)=1,F(2)=1,F(3)=-1$ and by identities F(2k)=F(k),F(2k+1)=F(k) for $ k>=2.$The sum $F(1)+F(2)+F(3)+...+F(100)$ is________

Comments

Yes ,in test series

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It is not correct question. leave it

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@srestha

see this https://undergroundmathematics.org/counting-and-binomials/r5875/solution

Answer 1

Let's start series by ignoring first 3 numbers. So F(4)=1,F(5)=1,F(6)=-1,F(7)=-1,F(8)=1,F(9)=1,F(10)=1,F(11)=1,

We observe that +1 and -1 will occur simultaneously as +2,-2,+4,-4,+8,-8,+16,-16,+32 .Now total terms are 92.So,until 97 5 times -1 will be there so,32 -5=27, Now adding +1 as F(0)+F(1)+F(2)=1 So,27+1=28

Comments

How you got this, So, until 97, 5 times -1 will be there so,32 -5=27?

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I think from $F(1)$ $to$ $F(95)=32+1=33$

and we find the$:F(96)=F(2.48)=F(48)=F(2.24)=F(24)=F(2.12)=F(12)=F(2.6)=F(6)=F(2.3)=F(3)=-1$

similarly we can find $F(97)=-1,F(98)=-1,F(99)=-1,F(100)=-1$

So,we get $F(1)+F(2)+...........+F(95)+F(96)+F(97)+F(98)+F(99)+F(100)$

$\Rightarrow33-1-1-1-1-1$

$\Rightarrow 28$

please correct me,if i'm wrong?

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After 32 +1 there will be 32 -1 .But as we have reached upto 92 (+3 as I stared from 4) already and we need to go upto 97( +3) as we will have only -5. So,32-5=27.

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Yes, see my comment is also right?