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Question Number 202780 by BaliramKumar last updated on 03/Jan/24

1+2+4+7+11+16+............up to 25 terms = ?

$$\mathrm{1}+\mathrm{2}+\mathrm{4}+\mathrm{7}+\mathrm{11}+\mathrm{16}+............\mathrm{up}\:\mathrm{to}\:\mathrm{25}\:\mathrm{terms}\:=\:? \\ $$

Answered by mr W last updated on 03/Jan/24

a_1 =1 a_2 =a_1 +1 a_3 =a_2 +2 a_4 =a_3 +3 ... a_n =a_(n−1) +(n−1) Σ: a_n =1+1+2+3+...+(n−1)=1+((n(n−1))/2) S_n =Σ_(k=1) ^n a_k =n+((n(n+1)(2n+1))/(12))−((n(n+1))/4) ⇒S_n =((n(n^2 +5))/6) with n=25: a_(25) =1+((25×24)/2)=301 S_(25) =((25(25^2 +5))/6)=2625 i.e. 1+2+4+7+11+16+...+301=2625

$${a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{\mathrm{2}} ={a}_{\mathrm{1}} +\mathrm{1} \\ $$$${a}_{\mathrm{3}} ={a}_{\mathrm{2}} +\mathrm{2} \\ $$$${a}_{\mathrm{4}} ={a}_{\mathrm{3}} +\mathrm{3} \\ $$$$... \\ $$$${a}_{{n}} ={a}_{{n}−\mathrm{1}} +\left({n}−\mathrm{1}\right) \\ $$$$\Sigma: \\ $$$${a}_{{n}} =\mathrm{1}+\mathrm{1}+\mathrm{2}+\mathrm{3}+...+\left({n}−\mathrm{1}\right)=\mathrm{1}+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{k}} ={n}+\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{12}}−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}} \\ $$$$\Rightarrow{S}_{{n}} =\frac{{n}\left({n}^{\mathrm{2}} +\mathrm{5}\right)}{\mathrm{6}} \\ $$$${with}\:{n}=\mathrm{25}: \\ $$$${a}_{\mathrm{25}} =\mathrm{1}+\frac{\mathrm{25}×\mathrm{24}}{\mathrm{2}}=\mathrm{301} \\ $$$${S}_{\mathrm{25}} =\frac{\mathrm{25}\left(\mathrm{25}^{\mathrm{2}} +\mathrm{5}\right)}{\mathrm{6}}=\mathrm{2625} \\ $$$${i}.{e}. \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{4}+\mathrm{7}+\mathrm{11}+\mathrm{16}+...+\mathrm{301}=\mathrm{2625} \\ $$

Commented by BaliramKumar last updated on 03/Jan/24

Thanks Sir

$$\mathrm{Thanks}\:\mathrm{Sir} \\ $$

Answered by talminator2856792 last updated on 03/Jan/24

1+2+3+4+...+n = ((n(n+1))/2) 1+3+6+10+...+ ((n(n+1))/2) = ((n(n+1)(n+2))/6) 1+2+4+7+11+...+a_n = (1+2+3+4+5+...+n) + (0+0+1+3+6+...+ (((n−1)(n−2))/2)) = ((n(n+1))/2) + ((n(n−1)(n−2))/6) → 1+2+4+7+11+16+...+a_(25) = 325+ 2300 = 2625

$$\:\:\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+...+{n} \\ $$$$\:\:=\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:\: \\ $$$$\:\:\mathrm{1}+\mathrm{3}+\mathrm{6}+\mathrm{10}+...+\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:\:=\:\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{6}} \\ $$$$\:\: \\ $$$$\:\:\mathrm{1}+\mathrm{2}+\mathrm{4}+\mathrm{7}+\mathrm{11}+...+{a}_{{n}} \\ $$$$\:\:=\:\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+...+{n}\right)\:+ \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{0}+\mathrm{0}+\mathrm{1}+\mathrm{3}+\mathrm{6}+...+\:\frac{\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{2}}\right) \\ $$$$\:\:=\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:+\:\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{6}} \\ $$$$\:\: \\ $$$$\:\:\rightarrow\:\mathrm{1}+\mathrm{2}+\mathrm{4}+\mathrm{7}+\mathrm{11}+\mathrm{16}+...+{a}_{\mathrm{25}} \\ $$$$\:\:=\:\mathrm{325}+\:\mathrm{2300} \\ $$$$\:\:=\:\mathrm{2625} \\ $$

Commented by BaliramKumar last updated on 03/Jan/24

Thanks Sir

$$\mathrm{Thanks}\:\mathrm{Sir} \\ $$

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