A) \[f(2n)-16f(n)\operatorname{for}\,al\operatorname{l}\,n\in N\]
B) \[f(n)-16f\left( \frac{n-1}{2} \right)\]when \[n\]is odd
C) \[f(n)-16f\left( \frac{n}{2} \right)\]when \[n\]is even
D) None of these
Correct Answer: A
Solution :
\[\sum\limits_{r=1}^{n}{{{(2r-1)}^{4}}={{1}^{4}}+{{3}^{4}}+{{5}^{4}}+.....+{{(2n-1)}^{4}}}\] |
\[={{1}^{4}}+{{2}^{4}}+{{3}^{4}}+....+{{(2n)}^{4}}-[{{2}^{4}}+{{4}^{4}}+....+{{(2n)}^{4}}]\] |
\[=f(2n)-16\{{{1}^{4}}+{{2}^{4}}+....+{{n}^{4}}\}\] |
\[=f(2n)-16f(n)\]for all \[n\in N;\] |
Whether \[n\]is even on odd. |
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