Putnam

Let a_1, a_2, . . . , a_n be real numbers, and let b_1, b_2, ..., b_n be distinct positive integers. Suppose that there is a polynomial f(x) satisfying the identity
(1 - x)^n f(x) = 1 + \sum _{i=1} ^n a_i x^{b_i}
Find the value of f(1) expressed in terms of b_1, b_2, ..., b_n and n .

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About shitikanth

I am a fourth year undergraduate student of Computer Science at IIT Kanpur. I love doing mathematics, and science. I also do a bit of coding in my free time.

10 responses to “Putnam”

  1. Jithendar says :

    Hoping to see more good problems here if you have any. Also, if I may ask, can you post some elegant solutions too.

  2. Ankit Kumar says :

    f(1) = 1 ans.

  3. SHUBHAM says :

    SHITIKANTH ISKA ANWER “0” HAI CHECK KRR LEH

  4. SHUBHAM says :

    VERY GOOD QUE AISEY AUR PLZZ ….

  5. SIVA NAGI REDDY says :

    Answer will not depend upon b1,b2…..bn’s. Answer is 1+a1+a2+…an (unless I made calculation errors).
    Take the nth roots of unity 1,w,w^2,…w^(n-1). Substitute into the equation and add all of them. We will get f(1)- constant term of f =a1+a2…+an. Since f(0)=1. f(1)=1+a1+a2+..+an

  6. Karthik says :

    Something like -b/n-2

    • Siva Modugula says :

      Differentiate both sides n times using the product rule and Leibniz formula, then x=1. We get f(1) = sum of (a_i.max(0, b_i.(b_i-1)……(b_i-n+1))) from i=1 to n.

  7. Yesh says :

    Edunovus is very helpful for slow learners like me.the vedios are very easy to understand.both the practicles and theory vedios are student friendly.

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