Question
Let $$\sum\limits_{n = 1}^n {{r^4} = f\left( n \right).} $$ Then $$\sum\limits_{r = 1}^n {{{\left( {2r - 1} \right)}^4}} $$ is equal to
A.
$$f\left( {2n} \right) - 16f\left( n \right)\,{\text{for all }}n \in N$$
B.
$$f\left( n \right) - 16f\left( {\frac{{n - 1}}{2}} \right)\,{\text{when }}n\,\,{\text{is odd}}$$
C.
$$f\left( n \right) - 16f\left( {\frac{{n}}{2}} \right)\,{\text{when }}n\,\,{\text{is even}}$$
D.
None of these
Answer :
$$f\left( {2n} \right) - 16f\left( n \right)\,{\text{for all }}n \in N$$
Solution :
$$\eqalign{
& \sum\limits_{r = 1}^n {{{\left( {2r - 1} \right)}^4} = {1^4} + {3^4} + {5^4} + ..... + {{\left( {2n - 1} \right)}^4}} \cr
& \sum\limits_{r = 1}^n {{{\left( {2r - 1} \right)}^4} = {1^4} + {2^4} + {3^4} + ..... + } {\left( {2n} \right)^4} - \left\{ {{2^4} + {4^4} + ..... + {{\left( {2n} \right)}^4}} \right\} \cr
& \sum\limits_{r = 1}^n {{{\left( {2r - 1} \right)}^4} = f\left( {2n} \right) - 16\left\{ {{1^4} + {2^4} + ..... + {n^4}} \right\} = f\left( {2n} \right) - 16f\left( n \right).} \cr
& {\text{If }}n = 2m,\,{\text{then }}\sum\limits_{r = 1}^n {{{\left( {2r - 1} \right)}^4} = {1^4} + {3^4} + {5^4} + ..... + {{\left( {4m - 1} \right)}^4}} \cr
& = {1^4} + {2^4} + {3^4} + ..... + {\left( {2m} \right)^4} + {\left( {2m + 1} \right)^4} + ..... + {\left( {4m - 1} \right)^4} + {\left( {4m} \right)^4} - \left\{ {{2^4} + {4^4} + ..... + {{\left( {4m} \right)}^4}} \right\} \cr
& = f\left( {4m} \right) - 16\left\{ {{1^4} + {2^4} + ..... + {{\left( {2m} \right)}^4}} \right\} \cr
& = f\left( {2n} \right) - 16f\left( n \right),{\text{e}}{\text{.t}}{\text{.c}}{\text{.}} \cr} $$