24will it have a diff ans ???
\sum_{i=0}^{\infty}{\sum_{j=0}^{\infty}{\sum_{k=0}^{\infty}{\frac{1}{3^i3^j3^k}}}}=??
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12 Answers
24Its not a dbt sir...........its for others...i know its very easy....not for kaymant sir,prophet sir ..and bhargav tooo [3][3]
62btw a far better one... (someone asked this to me on this saturday :D)
\huge \sum_{i=0}^{\infty}{\sum_{j=0}^{\infty}{\sum_{k=0;i\neq j,i\neq k,j\neq k}^{\infty}{\frac{1}{3^i3^j3^k}}}}
62yes ofcourse..
we are removing all the points where either i=j or i=k or j=k
1Sir...SOMEONE asked this to u this Saturday ????
Hmmmmmm......Who could it be ???? :-o :-)
1nishant bhaiya...i didnt get it...
this is to be solved using a step by step approach na? i mean we first solve the innermost Σ, then the one after that, and then the outermost...that means k takes all values b/w 0 and ∞...similarly for j and i...so how can we say that there are points for which i≠j, i≠k, j≠k?
21@shreyan take this 132 33 34 this is one such case wer i≠j, i≠k, j≠k
i m yet to do the second one which nishant sir gave
well giving the ans for the one which eureka posted
for i=0
13030[130 +......13∞]
+
13031[130 +......13∞]
.
.
.
1303∞[130 +......13∞]
__________________________________________________
for i=1
13130[130 +......13∞]
+
13131[130 +......13∞]
.
.
.
1313∞[130 +......13∞]
_________________________________________________.
.
.
.
.
.
.fro i=∞
13∞30[130 +......13∞]
+
13∞31[130 +......13∞]
.
.
.
.
13∞3∞[130 +......13∞]
now for i=0
we hav
summationa as
130[130 +......13∞] {130 +......13∞}
for i=1
we hav summation as
131[130 +......13∞] {130 +......13∞}
.
.
.
.
for i=∞
we hav summation as
13∞[130 +......13∞] {130 +......13∞}
now taking [130 ......13∞] {130 ......13∞} out we hav
=>>[130 ......13∞] {130 +......13∞}[{130 +......13∞}]
so ans is (3/2)(3/2)(3/2)=(3/2)3
24@eragon,
Its good to see that u made such a big effort..but it can be finished in jsut few lines
\sum_{i=0}^{\infty}{\sum_{j=0}^{\infty}{(\frac{1}{3^0}+\frac{1}{3^1}+...\infty)}}=\sum_{i=0}^{\infty}{\sum_{j=0}^{\infty}}(\frac{3}{2})
Just by this we can conclude that ans will be \frac{3}{2}.\frac{3}{2}.\frac{3}{2}=(\frac{3}{2})^3
21lol.....i dont think tat i made such a big effort....but ya initially i thought jus doing like tat....but then thought to explain it more properly bec some may say they din understand[3]
341far easier is to realize that it is just
\left(1+\frac{1}{3} + \frac{1}{3^2} +... \infty \right)\left(1+\frac{1}{3} + \frac{1}{3^2} +... \infty \right)\left(1+\frac{1}{3} + \frac{1}{3^2} +... \infty \right) = \left(\frac{3}{2} \right)^3