Question Number 166707 by amin96 last updated on 25/Feb/22 | ||
$$\frac{\mathrm{8}}{\mathrm{1}×\mathrm{5}×\mathrm{9}}+\frac{\mathrm{8}}{\mathrm{5}×\mathrm{9}×\mathrm{13}}+\frac{\mathrm{8}}{\mathrm{9}×\mathrm{13}×\mathrm{17}}+\ldots+\frac{\mathrm{1}}{\mathrm{41}×\mathrm{45}×\mathrm{49}}=? \\ $$$$ \\ $$$$ \\ $$$$\boldsymbol{{by}}\:\boldsymbol{{M}}.\boldsymbol{{A}} \\ $$ | ||
Answered by mahdipoor last updated on 25/Feb/22 | ||
$$\frac{\mathrm{8}}{{n}×\left({n}+\mathrm{4}\right)×\left({n}+\mathrm{8}\right)}=\frac{\mathrm{1}/\mathrm{4}}{{n}}−\frac{\mathrm{1}/\mathrm{2}}{{n}+\mathrm{4}}+\frac{\mathrm{1}/\mathrm{4}}{{n}+\mathrm{8}} \\ $$$$\Rightarrow\frac{\mathrm{8}}{\mathrm{1}×\mathrm{5}×\mathrm{9}}+\frac{\mathrm{8}}{\mathrm{5}×\mathrm{9}×\mathrm{13}}+\frac{\mathrm{8}}{\mathrm{9}×\mathrm{13}×\mathrm{17}}+... \\ $$$$+\frac{\mathrm{8}}{\mathrm{41}×\mathrm{45}×\mathrm{49}}= \\ $$$$\left(\frac{\mathrm{0}.\mathrm{25}}{\mathrm{1}}−\frac{\mathrm{0}.\mathrm{5}}{\mathrm{5}}+\frac{\mathrm{0}.\mathrm{25}}{\mathrm{9}}\right)+\left(\frac{\mathrm{0}.\mathrm{25}}{\mathrm{5}}−\frac{\mathrm{0}.\mathrm{5}}{\mathrm{9}}+\frac{\mathrm{0}.\mathrm{25}}{\mathrm{13}}\right)+ \\ $$$$+\left(\frac{\mathrm{0}.\mathrm{25}}{\mathrm{9}}−\frac{\mathrm{0}.\mathrm{5}}{\mathrm{13}}+\frac{\mathrm{0}.\mathrm{25}}{\mathrm{17}}\right)+...+\left(\frac{\mathrm{0}.\mathrm{25}}{\mathrm{41}}−\frac{\mathrm{0}.\mathrm{5}}{\mathrm{45}}+\frac{\mathrm{0}.\mathrm{25}}{\mathrm{49}}\right)= \\ $$$$=\left(\frac{\mathrm{0}.\mathrm{25}}{\mathrm{1}}−\frac{\mathrm{0}.\mathrm{5}}{\mathrm{5}}\right)+\left(\frac{\mathrm{0}.\mathrm{25}}{\mathrm{5}}\right)+\left(\frac{\mathrm{0}.\mathrm{25}}{\mathrm{45}}\right)+\left(−\frac{\mathrm{0}.\mathrm{5}}{\mathrm{45}}+\frac{\mathrm{0}.\mathrm{25}}{\mathrm{49}}\right)= \\ $$$$\mathrm{0}.\mathrm{2}−\frac{\mathrm{1}}{\mathrm{45}×\mathrm{49}} \\ $$$$ \\ $$ | ||