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Question Number 183064 by Matica last updated on 19/Dec/22

f(1)=1, f(n)=2Σ_(k=1) ^(n−1) f(k). find Σ_(k=1) ^(m) f(k).

$$\:{f}\left(\mathrm{1}\right)=\mathrm{1},\:\:{f}\left({n}\right)=\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\Sigma}}{f}\left({k}\right).\:\:{find}\:\underset{{k}=\mathrm{1}} {\overset{{m}} {\Sigma}}{f}\left({k}\right). \\ $$

Answered by mr W last updated on 19/Dec/22

a_n =f(n) s_n =Σ_(k=1) ^n a_k a_n =2(a_1 +a_2 +...+a_(n−1) ) 3a_n =2(a_1 +a_2 +...+a_(n−1) +a_n )=2s_n a_n =(2/3)s_n s_n −s_(n−1) =(2/3)s_n s_n =3s_(n−1) =3^2 s_(n−2) =3^(n−1) s_1 =3^(n−1) a_1 =3^(n−1) Σ_(k=1) ^m f(k)=s_m =3^(m−1) ✓

$${a}_{{n}} ={f}\left({n}\right) \\ $$$${s}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{k}} \\ $$$${a}_{{n}} =\mathrm{2}\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} +...+{a}_{{n}−\mathrm{1}} \right) \\ $$$$\mathrm{3}{a}_{{n}} =\mathrm{2}\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} +...+{a}_{{n}−\mathrm{1}} +{a}_{{n}} \right)=\mathrm{2}{s}_{{n}} \\ $$$${a}_{{n}} =\frac{\mathrm{2}}{\mathrm{3}}{s}_{{n}} \\ $$$${s}_{{n}} −{s}_{{n}−\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}}{s}_{{n}} \\ $$$${s}_{{n}} =\mathrm{3}{s}_{{n}−\mathrm{1}} =\mathrm{3}^{\mathrm{2}} {s}_{{n}−\mathrm{2}} =\mathrm{3}^{{n}−\mathrm{1}} {s}_{\mathrm{1}} =\mathrm{3}^{{n}−\mathrm{1}} {a}_{\mathrm{1}} =\mathrm{3}^{{n}−\mathrm{1}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}{f}\left({k}\right)={s}_{{m}} =\mathrm{3}^{{m}−\mathrm{1}} \:\checkmark \\ $$

Commented by Matica last updated on 19/Dec/22

Thank a lot.

$${Thank}\:{a}\:{lot}.\: \\ $$

Commented by Matica last updated on 19/Dec/22

How did you get s_1 =a_1 ? why not a_1 =(2/3)s_1 ?

$${How}\:{did}\:{you}\:{get}\:{s}_{\mathrm{1}} ={a}_{\mathrm{1}} ?\:{why}\:{not}\:{a}_{\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}}{s}_{\mathrm{1}} ? \\ $$

Commented by mr W last updated on 19/Dec/22

s_n means the sum of first n terms. a_1 =1, s_1 =a_1 =1 these are given initial conditions. s_1 is the sum of the first term, it must be a_1 ! or do you think a_1 =(2/3)s_1 makes any sense?

$${s}_{{n}} \:{means}\:{the}\:{sum}\:{of}\:{first}\:{n}\:{terms}. \\ $$$${a}_{\mathrm{1}} =\mathrm{1},\:{s}_{\mathrm{1}} ={a}_{\mathrm{1}} =\mathrm{1} \\ $$$${these}\:{are}\:{given}\:{initial}\:{conditions}. \\ $$$${s}_{\mathrm{1}} \:{is}\:{the}\:{sum}\:{of}\:{the}\:{first}\:{term},\:{it}\:{must} \\ $$$${be}\:{a}_{\mathrm{1}} !\:{or}\:{do}\:{you}\:{think}\:{a}_{\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}}{s}_{\mathrm{1}} \:{makes} \\ $$$${any}\:{sense}? \\ $$

Commented by Matica last updated on 19/Dec/22

Ok, sir. i decided.

$${Ok},\:{sir}.\:{i}\:{decided}. \\ $$

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