Suppose for each n∈N,14+24+34+…+n4=an5+bn4+cn3+dn2+en+f, then value of b is 

Suppose for each nN,14+24+34++n4=an5+bn4+cn3+dn2+en+f, then value of b is 

  1. A

    1/5

  2. B

    1/2

  3. C

    1/3

  4. D

    1

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    Solution:

    We have 
    14+24++n4=an5+bn4+cn3+dn2+en+f          (1)
    and 14+24++n4+(n+1)4
           =a(n+1)5+b(n+1)4+c(n+1)3+d(n+1)2+e(n+1)+f         (2)
    Subtracting (1) from (2), we get
    (n+1)4=a(n+1)5n5+b(n+1)4n4                 +c(n+1)3n3+d(n+1)2n2                 +e[(n+1)n]
    Comparing coefficients fo n4 , we get
                1=5a  a=1/5
    Comparing coefficient fo n3 , we get
              4=a 5C2+4bb=1/2.

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