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let (1^4+2^4+3^4+....n^4)=f(n).then the sum of ∑ (2r-1)^4=?(where r=1.....n) f(2n)-16f(n ) f(2n)-f(n) f(2n)-4f(n) noe of these

let (1^4+2^4+3^4+....n^4)=f(n).then the sum of
(2r-1)^4=?(where r=1.....n)
  1. f(2n)-16f(n)
  2. f(2n)-f(n)
  3. f(2n)-4f(n)
  4. noe of these

Grade:12

1 Answers

Sourabh Singh IIT Patna
askIITians Faculty 2104 Points
7 years ago
use the summation terms given in this equation
(1^4+2^4+3^4+....n^4)=f(n)
and the given summation to find can be seen as the sum of odd terms of (2n-1) terms.First try to write f(2n) an dthen take out common of 24 from even terms and you will be left with the summation u have to find.
After a very simple reaarangement you will see that you will get the correct answer as A

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