Question Number 202780 by BaliramKumar last updated on 03/Jan/24
$$\mathrm{1}+\mathrm{2}+\mathrm{4}+\mathrm{7}+\mathrm{11}+\mathrm{16}+…………\mathrm{up}\:\mathrm{to}\:\mathrm{25}\:\mathrm{terms}\:=\:? \\ $$
Answered by mr W last updated on 03/Jan/24
$${a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{\mathrm{2}} ={a}_{\mathrm{1}} +\mathrm{1} \\ $$$${a}_{\mathrm{3}} ={a}_{\mathrm{2}} +\mathrm{2} \\ $$$${a}_{\mathrm{4}} ={a}_{\mathrm{3}} +\mathrm{3} \\ $$$$… \\ $$$${a}_{{n}} ={a}_{{n}−\mathrm{1}} +\left({n}−\mathrm{1}\right) \\ $$$$\Sigma: \\ $$$${a}_{{n}} =\mathrm{1}+\mathrm{1}+\mathrm{2}+\mathrm{3}+…+\left({n}−\mathrm{1}\right)=\mathrm{1}+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{k}} ={n}+\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{12}}−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}} \\ $$$$\Rightarrow{S}_{{n}} =\frac{{n}\left({n}^{\mathrm{2}} +\mathrm{5}\right)}{\mathrm{6}} \\ $$$${with}\:{n}=\mathrm{25}: \\ $$$${a}_{\mathrm{25}} =\mathrm{1}+\frac{\mathrm{25}×\mathrm{24}}{\mathrm{2}}=\mathrm{301} \\ $$$${S}_{\mathrm{25}} =\frac{\mathrm{25}\left(\mathrm{25}^{\mathrm{2}} +\mathrm{5}\right)}{\mathrm{6}}=\mathrm{2625} \\ $$$${i}.{e}. \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{4}+\mathrm{7}+\mathrm{11}+\mathrm{16}+…+\mathrm{301}=\mathrm{2625} \\ $$
Commented by BaliramKumar last updated on 03/Jan/24
$$\mathrm{Thanks}\:\mathrm{Sir} \\ $$
Answered by talminator2856792 last updated on 03/Jan/24
$$\:\:\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+…+{n} \\ $$$$\:\:=\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:\: \\ $$$$\:\:\mathrm{1}+\mathrm{3}+\mathrm{6}+\mathrm{10}+…+\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:\:=\:\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{6}} \\ $$$$\:\: \\ $$$$\:\:\mathrm{1}+\mathrm{2}+\mathrm{4}+\mathrm{7}+\mathrm{11}+…+{a}_{{n}} \\ $$$$\:\:=\:\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+…+{n}\right)\:+ \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{0}+\mathrm{0}+\mathrm{1}+\mathrm{3}+\mathrm{6}+…+\:\frac{\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{2}}\right) \\ $$$$\:\:=\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:+\:\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{6}} \\ $$$$\:\: \\ $$$$\:\:\rightarrow\:\mathrm{1}+\mathrm{2}+\mathrm{4}+\mathrm{7}+\mathrm{11}+\mathrm{16}+…+{a}_{\mathrm{25}} \\ $$$$\:\:=\:\mathrm{325}+\:\mathrm{2300} \\ $$$$\:\:=\:\mathrm{2625} \\ $$
Commented by BaliramKumar last updated on 03/Jan/24
$$\mathrm{Thanks}\:\mathrm{Sir} \\ $$