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Question-84783




Question Number 84783 by Power last updated on 16/Mar/20
Commented by Power last updated on 16/Mar/20
k=3
$$\mathrm{k}=\mathrm{3} \\ $$
Commented by Tony Lin last updated on 16/Mar/20
what does k=3 mean?
$${what}\:{does}\:{k}=\mathrm{3}\:{mean}? \\ $$
Commented by Power last updated on 16/Mar/20
k≠1  ⇒ k=3    instaed of k=1
$$\mathrm{k}\neq\mathrm{1}\:\:\Rightarrow\:\mathrm{k}=\mathrm{3}\:\:\:\:\mathrm{instaed}\:\mathrm{of}\:\mathrm{k}=\mathrm{1} \\ $$
Commented by Power last updated on 16/Mar/20
please see
$$\mathrm{please}\:\mathrm{see} \\ $$
Commented by Power last updated on 16/Mar/20
sir MJS pls
$$\mathrm{sir}\:\mathrm{MJS}\:\mathrm{pls} \\ $$
Commented by Tony Lin last updated on 16/Mar/20
i don′t know the quick way to evaluate  the answer  but the answer is very ugly  ((364219040265251)/(13127595717600))  ≒27.7445351  it seems that there is no quick way   to solve  just add altogether
$${i}\:{don}'{t}\:{know}\:{the}\:{quick}\:{way}\:{to}\:{evaluate} \\ $$$${the}\:{answer} \\ $$$${but}\:{the}\:{answer}\:{is}\:{very}\:{ugly} \\ $$$$\frac{\mathrm{364219040265251}}{\mathrm{13127595717600}} \\ $$$$\fallingdotseq\mathrm{27}.\mathrm{7445351} \\ $$$${it}\:{seems}\:{that}\:{there}\:{is}\:{no}\:{quick}\:{way}\: \\ $$$${to}\:{solve} \\ $$$${just}\:{add}\:{altogether} \\ $$
Answered by mind is power last updated on 16/Mar/20
=Σ_(k=1) ^(32) (1−(1/(k+1)))  =32−Σ_(k=1) ^(32) (1/(k+1))=32−Σ_(k=2) ^(33) (1/k)=33−H_(33)
$$=\underset{{k}=\mathrm{1}} {\overset{\mathrm{32}} {\sum}}\left(\mathrm{1}−\frac{\mathrm{1}}{{k}+\mathrm{1}}\right) \\ $$$$=\mathrm{32}−\underset{{k}=\mathrm{1}} {\overset{\mathrm{32}} {\sum}}\frac{\mathrm{1}}{{k}+\mathrm{1}}=\mathrm{32}−\underset{{k}=\mathrm{2}} {\overset{\mathrm{33}} {\sum}}\frac{\mathrm{1}}{{k}}=\mathrm{33}−{H}_{\mathrm{33}} \\ $$
Commented by Power last updated on 16/Mar/20
thanks   H_(33)   −?
$$\mathrm{thanks}\:\:\:\mathrm{H}_{\mathrm{33}} \:\:−? \\ $$
Commented by mind is power last updated on 16/Mar/20
H_n =Σ_(k=1) ^n (1/k)
$${H}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}} \\ $$

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