Question Number 107609 by mathdave last updated on 11/Aug/20 | ||
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Answered by mr W last updated on 11/Aug/20 | ||
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$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}+\mathrm{4}} \\ $$$$=\underset{{k}=\mathrm{5}} {\overset{{n}+\mathrm{4}} {\sum}}\frac{\mathrm{1}}{{k}} \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}+\mathrm{4}} {\sum}}\frac{\mathrm{1}}{{k}}−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$={H}_{{n}+\mathrm{4}} −\frac{\mathrm{25}}{\mathrm{12}} \\ $$ | ||
Commented by mathdave last updated on 11/Aug/20 | ||
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$${i}\:{really}\:{appreciated}\:{ds}\:\:{thankx} \\ $$ | ||
Answered by hgrocks last updated on 11/Aug/20 | ||
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$$\mathrm{S}_{\mathrm{n}} =\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}+\mathrm{x}}\:=\:\:\underset{\mathrm{k}=\mathrm{1}+\mathrm{x}} {\overset{\mathrm{n}+\mathrm{x}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}}\:=\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}+\mathrm{x}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}}\:−\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{x}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}} \\ $$$$\:\mathrm{S}_{\mathrm{n}} \:\:\:\:=\:\mathrm{H}_{\mathrm{n}+\mathrm{x}} \:−\:\mathrm{H}_{\mathrm{x}} \\ $$$$ \\ $$$$ \\ $$ | ||
Commented by mathdave last updated on 11/Aug/20 | ||
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$${thank}\:{so}\:{so}\:{much} \\ $$ | ||