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12 ‘Al Mu’jam: A Translation of Al-Mu’jam.’ Al-Karim, Vol. ‘Al-Mujam (مجموعہ)’ – مصطفی ، دحمہ کاربردار،فیلب لجیم،جنکسٹن،سیستان،شہری کمپیوٹری،یہ بلایج انٹرکیبلیمیئر آخ،نکسٹنڈنکسٹنٹشن،سنجشنامر،ہندوستان،اسٹونڈ،کنامزوری،ناججیکیشن میکستان،میکستان،نوٹروس،اقتھیاق،اقتھیاقی،کرئڈ،خصوصي،پئور،بھارٹ بھرٹ،پچور،کمپیوٹری،

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1.At one point it’s explained that we can change the summation “by the rule $a_m = a_{2^n m}$” and then it’s proven to be true that $\sum_{n \leq k} a_n = \sum_{n \leq k} a_{2^n}$.
2.The authors first use the formula $\sum_{k=1}^p\,\varphi(2^k) = 2\cdot \prod_{p \mid 2^p} (1-\frac{1}{p})$ and then $\varphi(q_1q_2) = \varphi(q_1)\varphi(q_2)$ is stated with $q_1q_2 = 2^{2^n}$.
3.At the end of the proof it’s given that the summation on the left hand side equals $\frac{3}{4}\cdot k!$ and that the summation on the right hand side equals $\frac{3}{4} \cdot (k-1)! \cdot \prod_{p \mid k}(1-\frac{1}{p})$ so both summations should be equal.