[tex]\displaystyle\sum^{n}_{k=1}(2k-1)a_{k}=n(n+1)(4n-1)[/tex]
Find [tex]a_{n}[/tex].
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[tex]\displaystyle\sum^{n}_{k=1}(2k-1)a_{k}=n(n+1)(4n-1)[/tex]
Find [tex]a_{n}[/tex].
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We have,
[tex]\rm{\displaystyle\sum^{n}_{k=1}(2k-1)a_{n}=n(n+1)(4n-1)}[/tex]
[tex]\rm{\displaystyle\sum^{n+1}_{k=1}(2k-1)a_{n}=(n+1)(n+2)(4n+3)}[/tex]
Subtracting,
[tex]\small\text{$\rm{\displaystyle\sum^{n+1}_{k=1}(2k-1)a_{n}-\displaystyle\sum^{n}_{k=1}(2k-1)a_{n}=(n+1)(n+2)(4n+3)-n(n+1)(4n-1)}$}[/tex]
[tex]\rm{(2n+1)a_{n+1}=(n+1)(4n^{2}+11n+6-4n^{2}+n)}[/tex]
[tex]\rm{(2n+1)a_{n+1}=(n+1)(12n+6)}[/tex]
[tex]\rm{a_{n+1}=6(n+1)}[/tex]
[tex]\rm{\therefore a_{n}=6n}[/tex]
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