Find all positive integers n for which there exist non−negative integer . a_(1 ) a_2 a_3 ....... a_n . Such that (1/2^a_1 ) + (1/2^a_2 ) + (1/2^a_3 ) + .... + (1/2^a_n ) = (1/3^a_1 ) + (2/3^a_2 ) + .... + (n/3^a


Question and Answers Forum

All Questions Topic List
Arithmetic Questions
Previous in All Question Next in All Question
Previous in Arithmetic Next in Arithmetic
Question Number 5834 by sanusihammed last updated on 31/May/16

$F i n d a l l p o s i t i v e i n t e g e r s n f o r w h i c h t h e r e e x i s t$ $n o n - n e g a t i v e i n t e g e r . a_{1} a_{2} a_{3} . . . . . . . a_{n} . S u c h t h a t$ $\frac{1}{2^{a_{1}}} + \frac{1}{2^{a_{2}}} + \frac{1}{2^{a_{3}}} + . . . . + \frac{1}{2^{a_{n}}} = \frac{1}{3^{a_{1}}} + \frac{2}{3^{a_{2}}} + . . . . + \frac{n}{3^{a_{n}}} = 1$ $P l e a s e h e l p .$
Commented by Yozzii last updated on 31/May/16
$n=1⇒(1/2^a_1 )=(1/3^a_1 )⇒a_1 =0. Σ_(i=1) ^n (1/2^a_i )=Σ_(i=1) ^n (i/3^a_o ) Σ_(i=1) ^n ((1/2^a_i )−(i/3^a_i ))=0 Σ_(i=1) ^n (((3^a_i −i2^a_i )/6^a_i ))=0 6^(Σ_(l=1) ^n a_l ) (Σ_(i=1) ^n ((3^a_i −i2^a_i )/(6a^i )))=0 Σ_(i=1) ^n {6^(Σ_(l=1) ^n a_l −a_i ) (3^a_i −i2^a_i )}=0$
$n = 1 \Rightarrow \frac{1}{2^{a_{1}}} = \frac{1}{3^{a_{1}}} \Rightarrow a_{1} = 0 .$ $\underset{i = 1}{\sum^{n}} \frac{1}{2^{a_{i}}} = \underset{i = 1}{\sum^{n}} \frac{i}{3^{a_{o}}}$ $\underset{i = 1}{\sum^{n}} (\frac{1}{2^{a_{i}}} - \frac{i}{3^{a_{i}}}) = 0$ $\underset{i = 1}{\sum^{n}} (\frac{3^{a_{i}} - i 2^{a_{i}}}{6^{a_{i}}}) = 0$ $6^{\underset{l = 1}{\sum^{n}} a_{l}} (\underset{i = 1}{\sum^{n}} \frac{3^{a_{i}} - i 2^{a_{i}}}{6 a^{i}}) = 0$ $\underset{i = 1}{\sum^{n}} {6^{\underset{l = 1}{\sum^{n}} a_{l} - a_{i}} (3^{a_{i}} - i 2^{a_{i}})} = 0$
Commented by sanusihammed last updated on 31/May/16

$I a p p r e c i a t e$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum