Question Number 204348 by BaliramKumar last updated on 14/Feb/24
$$\mathrm{3}×\mathrm{7}×\mathrm{11}\:+\:\mathrm{7}×\mathrm{11}×\mathrm{15}\:+\:\mathrm{11}×\mathrm{15}×\mathrm{19}\:+\:……….+\:\:\mathrm{39}×\mathrm{43}×\mathrm{47}\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Answered by MM42 last updated on 14/Feb/24
$$\underset{\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\:\left(\mathrm{4}{n}−\mathrm{1}\right)\left(\mathrm{4}{n}+\mathrm{3}\right)\left(\mathrm{4}{n}+\mathrm{7}\right) \\ $$$$=\underset{\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\left(\mathrm{64}{n}^{\mathrm{3}} +\mathrm{14}{n}^{\mathrm{2}} +\mathrm{44}{n}−\mathrm{21}\right) \\ $$$$=\mathrm{64}×\left(\frac{\mathrm{10}×\mathrm{11}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{14}×\frac{\mathrm{10}×\mathrm{11}×\mathrm{21}}{\mathrm{6}}+\mathrm{44}×\frac{\mathrm{10}×\mathrm{11}}{\mathrm{2}}−\mathrm{210} \\ $$$$ \\ $$
Commented by BaliramKumar last updated on 14/Feb/24
$$\mathrm{thanks}\:\mathrm{sir} \\ $$$$\mathrm{typo}\:\:\:\:\:\mathrm{14}\rightarrow\mathrm{144} \\ $$$$\mathrm{also}\:\:\:\frac{\mathrm{39}×\mathrm{43}×\mathrm{47}×\left(\mathrm{47}\:+\:\mathrm{d}\right)}{\mathrm{d}\centerdot\mathrm{t}}\:−\:\frac{\left(\mathrm{3}\:−\:\mathrm{d}\right)×\mathrm{3}×\mathrm{7}×\mathrm{11}}{\mathrm{d}\centerdot\mathrm{t}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{d}\:=\:\mathrm{difference}\:=\:\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{t}\:=\:\mathrm{no}.\:\mathrm{of}\:\mathrm{terms}\:=\:\mathrm{3}\:+\:\mathrm{1}\:=\:\mathrm{4} \\ $$$$ \\ $$