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Question-149189

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Question Number 149189 by Naser last updated on 03/Aug/21
Commented by Tawa11 last updated on 03/Aug/21
great
$$\mathrm{great} \\ $$
Answered by Ar Brandon last updated on 03/Aug/21
S=Σ_(k=1) ^n (1/((3k−2)(2k+1))) =Σ_(k=1) ^n ((3/7)∙(1/((3k−2)))−(2/7)∙(1/((2k+1)))) =(1/7)Σ_(k=1) ^n (1/(k−(2/3)))−(1/(k+(1/2)))
$${S}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{k}−\mathrm{2}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)} \\ $$$$\:\:\:\:=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{3}}{\mathrm{7}}\centerdot\frac{\mathrm{1}}{\left(\mathrm{3}{k}−\mathrm{2}\right)}−\frac{\mathrm{2}}{\mathrm{7}}\centerdot\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{7}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}−\frac{\mathrm{2}}{\mathrm{3}}}−\frac{\mathrm{1}}{{k}+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$
Answered by nimnim last updated on 03/Aug/21
Let me give a try. S=Σ_(k=1) ^n (1/((3k−2)(2k+1))) =Σ_(k=1) ^n ((3/7).(1/(3k−2))−(2/7).(1/(2k+1))) =(3/7)Σ_(k=1) ^n (1/((3k−2)))−(2/7)Σ_(k=1) ^n (1/((2k+1))) =(3/7)((1/1)+(1/4)+(1/7)+....+(1/(1+(n−1).3))) −(2/7)((1/3)+(1/5)+(1/7)+...+(1/(3+(n−1).2))) both are harmonic progression to n terms S=(1/d)ln∣(((2a+(2n−1)d)/(2a−d)))∣ ∴S=(3/7)×(1/3)ln∣(((2.1+(2n−1).3)/(2.1−3)))∣−(2/7)×(1/2)ln∣(((2.3+(2n−1).2)/(2.3−2)))∣ =(1/7)ln∣((2n−1)/(−1))∣−(1/7)ln∣((4+4n)/4)∣ =(1/7)(ln∣1−2n∣−ln∣1+n∣) =(1/7)ln∣((1−2n)/(1+n))∣
$${Let}\:{me}\:{give}\:{a}\:{try}. \\ $$$${S}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{k}−\mathrm{2}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)} \\ $$$$\:\:\:=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{3}}{\mathrm{7}}.\frac{\mathrm{1}}{\mathrm{3}{k}−\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{7}}.\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\right) \\ $$$$\:\:\:=\frac{\mathrm{3}}{\mathrm{7}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{\left(\mathrm{3}{k}−\mathrm{2}\right)}−\frac{\mathrm{2}}{\mathrm{7}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)} \\ $$$$\:\:\:=\frac{\mathrm{3}}{\mathrm{7}}\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{7}}+….+\frac{\mathrm{1}}{\mathrm{1}+\left({n}−\mathrm{1}\right).\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{2}}{\mathrm{7}}\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{7}}+…+\frac{\mathrm{1}}{\mathrm{3}+\left({n}−\mathrm{1}\right).\mathrm{2}}\right) \\ $$$$\:\:\:\:\:{both}\:{are}\:{harmonic}\:{progression}\:{to}\:{n}\:{terms} \\ $$$$\:\:\:\:\:\:{S}=\frac{\mathrm{1}}{{d}}{ln}\mid\left(\frac{\mathrm{2}{a}+\left(\mathrm{2}{n}−\mathrm{1}\right){d}}{\mathrm{2}{a}−{d}}\right)\mid \\ $$$$\therefore{S}=\frac{\mathrm{3}}{\mathrm{7}}×\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid\left(\frac{\mathrm{2}.\mathrm{1}+\left(\mathrm{2}{n}−\mathrm{1}\right).\mathrm{3}}{\mathrm{2}.\mathrm{1}−\mathrm{3}}\right)\mid−\frac{\mathrm{2}}{\mathrm{7}}×\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\left(\frac{\mathrm{2}.\mathrm{3}+\left(\mathrm{2}{n}−\mathrm{1}\right).\mathrm{2}}{\mathrm{2}.\mathrm{3}−\mathrm{2}}\right)\mid \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{7}}{ln}\mid\frac{\mathrm{2}{n}−\mathrm{1}}{−\mathrm{1}}\mid−\frac{\mathrm{1}}{\mathrm{7}}{ln}\mid\frac{\mathrm{4}+\mathrm{4}{n}}{\mathrm{4}}\mid \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{7}}\left({ln}\mid\mathrm{1}−\mathrm{2}{n}\mid−{ln}\mid\mathrm{1}+{n}\mid\right) \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{7}}{ln}\mid\frac{\mathrm{1}−\mathrm{2}{n}}{\mathrm{1}+{n}}\mid \\ $$$$\:\:\:\:\:\: \\ $$

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