Question Number 144553 by mathdanisur last updated on 26/Jun/21
$${Solve}\:{for}\:{natural}\:{numbers}: \\ $$$$\frac{\mathrm{1}^{\mathrm{4}} }{\boldsymbol{{z}}}\:+\:\frac{\mathrm{2}^{\mathrm{4}} }{\boldsymbol{{z}}+\mathrm{1}}\:+\:\frac{\mathrm{3}^{\mathrm{4}} }{\boldsymbol{{z}}+\mathrm{2}}\:+\:…\:+\:\frac{\mathrm{10}^{\mathrm{4}} }{\boldsymbol{{z}}+\mathrm{9}}\:=\:\mathrm{3025} \\ $$
Answered by Olaf_Thorendsen last updated on 26/Jun/21
$$\mathrm{S}\left({z}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\mathrm{9}} {\sum}}\frac{\left({n}+\mathrm{1}\right)^{\mathrm{4}} }{{z}+{n}} \\ $$$$\mathrm{S}\left(\mathrm{1}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\mathrm{9}} {\sum}}\frac{\left({n}+\mathrm{1}\right)^{\mathrm{4}} }{{n}+\mathrm{1}}\:=\:\underset{{n}=\mathrm{0}} {\overset{\mathrm{9}} {\sum}}\left({n}+\mathrm{1}\right)^{\mathrm{3}} \:=\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}{n}^{\mathrm{3}} \\ $$$$\mathrm{S}\left(\mathrm{1}\right)\:=\:\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\mid_{{n}=\mathrm{10}} \:=\:\frac{\mathrm{100}×\mathrm{121}}{\mathrm{4}}\:=\:\mathrm{3025} \\ $$$$\Rightarrow\:\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}. \\ $$$$\mathrm{if}\:{z}\:\:\geqslant\:\mathrm{2},\:\mathrm{S}\left({z}\right)\:\leqslant\:\mathrm{S}\left(\mathrm{1}\right)\:=\:\mathrm{3025}\::\:\mathrm{no}\:\mathrm{solution} \\ $$
Commented by mathdanisur last updated on 26/Jun/21
$${cool}\:{thank}\:{you}\:{Sir} \\ $$