71Thanks! (for labour and especially that special mention)
18 Answers
71@kunl, Yes I too have ab bit problem with opening / analysing big summation expressions.
Ricky Bhaiya, Can you help me out with that!
71
49Now, addressing the original type of sum...
S = \sum_{i=1}^{m}{\sum_{j=1}^{m}{\sum_{k=1}^{m}{\frac{1}{3^i.3^j.3^k}}}}
If you have a good imagination in 3-d you can always imagine a "Three dimensional matrix" like the two dimensional matrix we had for the previous case...
We will have to sum up things like..
S = \frac{1}{3^1.3^1.3^1} + \frac{1}{3^1.3^1.3^2} + \frac{1}{3^1.3^1.3^3} ... + \frac{1}{3^1.3^2.3^1} + \frac{1}{3^1.3^2.3^2} + \frac{1}{3^1.3^2.3^3} ... + \frac{1}{3^1.3^3.3^1} + \frac{1}{3^1.3^3.3^2} + \frac{1}{3^1.3^3.3^3} ... + ... ... ... + \frac{1}{3^2.3^1.3^1} + \frac{1}{3^2.3^1.3^2} + \frac{1}{3^2.3^1.3^3} ... + \frac{1}{3^2.3^2.3^1} + \frac{1}{3^2.3^2.3^2} + \frac{1}{3^2.3^2.3^3} ... + \frac{1}{3^2.3^3.3^1} + \frac{1}{3^2.3^3.3^2} + \frac{1}{3^2.3^3.3^3} ... + ... ... ... + \frac{1}{3^3.3^1.3^1} + \frac{1}{3^3.3^1.3^2} + \frac{1}{3^3.3^1.3^3} ... + \frac{1}{3^3.3^2.3^1} + \frac{1}{3^3.3^2.3^2} + \frac{1}{3^3.3^2.3^3} ... + \frac{1}{3^3.3^3.3^1} + \frac{1}{3^3.3^3.3^2} + \frac{1}{3^3.3^3.3^3} ... + ... ... ...
That is,
S = \left[\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+... \right]\left[\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+... \right]\left[\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+... \right]
Another easy infinite g.p.
Now the sum given has i\neq j\neq k
i.e. none of i,j or k can have the same value at the same time...
Thus the sum will be less than the sum with no restriction no doubt..
But how much less..??
we will have to subtract the cases in which any 2 of i,j or k is equal and the case where all the 3 are equal...
S= sum with no restriction - 3C2.(sum when i=j)
now if we see, we can observe tht we have already subtracted the i=j=k case thrice...so to make up we will add up the i=j=k case twice and get desired answer! :)
S = sum with no restriction - 3C2.(sum when i=j) + 2.(sum when i=j=k)
P.S.:Practice of sums with sigma will make sigma less of a jaw coming to bite off ur head and more of a mere easy and obvious mathematical sign like '+',\; '-',\; 'x'\; or\; '\div '
I posted the above things as Vivek asked a detailed discussion
49Suppose the sum is:
S = \sum_{i=1}^{m}{\sum_{j=1}^{m}{\frac{1}{3^i.3^j}}} \; ; \; i\geq j;\; m\epsilon I^+
Then, we have to find out the sum of the elements of the m X m matrix excepting the elements that lie below the leading diagonal (why?)
Again if the summation is :
S = \sum_{i=1}^{m}{\sum_{j=1}^{m}{\frac{1}{3^i.3^j}}} \; ; \; j\geq i;\; m\epsilon I^+
Then, we have to find out the sum of the elements of the m X m matrix excepting the elements that lie above the leading diagonal (why?)
Now a bit more practice will make people the master in such sums...
49Suppose we have to find the summation:
S = \sum_{i=1}^{m}{\sum_{j=1}^{m}{\frac{1}{3^i.3^j}}} \; ; \; i\neq j;\; m\epsilon I^+
Then we have to find the sum of the elements of the mXm matrix excepting the diagonal elements...
\begin{bmatrix} \frac{1}{3^1.3^1} & \frac{1}{3^1.3^2} & \frac{1}{3^1.3^3} & \frac{1}{3^1.3^4} & ...\\ \frac{1}{3^2.3^1} & \frac{1}{3^2.3^2} & \frac{1}{3^2.3^3} & \frac{1}{3^2.3^4} & ...\\ \frac{1}{3^3.3^1} & \frac{1}{3^3.3^2} & \frac{1}{3^3.3^3} & \frac{1}{3^3.3^4} & ...\\ \frac{1}{3^4.3^1} & \frac{1}{3^4.3^2} & \frac{1}{3^4.3^3} & \frac{1}{3^4.3^4} & ...\\ ... & ... & ... & ... & ... \end{bmatrix}
Thus we can also write mathematically:
S = \left[\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}... \right].\left[\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}... \right] - \sum_{i=0}^{m}{\frac{1}{3^i.3^i}}
Which are easy infinite G.P.s waiting to be evaluated..!
49now lets take this to the next level....
suppose sum is:
S = \sum_{i=1}^{\propto}{\sum_{j=1}^{\propto}{\frac{1}{3^i.3^j}}}
This would mean...
S = \frac{1}{3^1.3^1} + \frac{1}{3^1.3^2} + \frac{1}{3^1.3^3} + ... ... + \frac{1}{3^2.3^1}+\frac{1}{3^2.3^2} + \frac{1}{3^2.3^3} + ... ...\frac{1}{3^3.3^1} + \frac{1}{3^3.3^2} + \frac{1}{3^3.3^3} + ... ... + \frac{1}{3^4.3^1}+\frac{1}{3^4.3^2} + \frac{1}{3^4.3^3} + ... ...
Thus as we can see,
S = \left[\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}... \right].\left[\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}... \right]
We can also sum up the summation as sum of all the elements of the matrix...where i,j element is signified by \frac{1}{3^i.3^j}
i.e. \begin{bmatrix} \frac{1}{3^1.3^1} & \frac{1}{3^1.3^2} & \frac{1}{3^1.3^3} & \frac{1}{3^1.3^4} & ...\\ \frac{1}{3^2.3^1} & \frac{1}{3^2.3^2} & \frac{1}{3^2.3^3} & \frac{1}{3^2.3^4} & ...\\ \frac{1}{3^3.3^1} & \frac{1}{3^3.3^2} & \frac{1}{3^3.3^3} & \frac{1}{3^3.3^4} & ...\\ \frac{1}{3^4.3^1} & \frac{1}{3^4.3^2} & \frac{1}{3^4.3^3} & \frac{1}{3^4.3^4} & ...\\ ... & ... & ... & ... & ... \end{bmatrix}
Next post comes in a few minutes...
49Chal ab jo JEE khatam (not too well tho) I can waste my time latex-ifying the soln.! :P
dekho it is something like this...
if the sum is...
S\; =\sum_{i=1}^{\propto}{\frac{1}{3^i}}
as u wud know it means...
S = \frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}....
easy convergent infinite GP...
next post comes in few mins..
1@ricky
to be very frank i don't feel ashamed to admit....that i did not even get what those horrible looking "three" sigmas mean and what are the terms.infact i did not get wat a typical term would be....so it would be great if u would explain wat does that expression really mean?
1Let us define a few terms below : -
Instead of directly summing the given expression , we shall try to find its value by omitting the sum of the unfavourable cases ( " S1 " and " S2 " ) from the total sum ( " S3 " ) .
Now , we can rewrite and sum up " S1 " as shown : -
Or , 
Similarly , we can also rewrite " S2 " and sum it up as : -
Or , 
Now , if one closely inspects " S2 " , he / she would realise that we have to subtract it 3 times from the total sum , that is " S3 " , since there are the following 3 cases : -
i ≠j = k ; i = j ≠k ; i = k ≠j .
Consequently , we compute the sum as follows : -
1(\frac{1}{1-\frac{1}{3}})^3 = \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\frac{1}{3^i3^j3^k}(i \neq j\neq k)+\sum_{i=0}^{\infty}{\frac{1}{3^{3i}}}+\underline{\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\frac{1}{3^{2i}3^j}} \\ \\ \text{Now the problem boils down in finding the underlined sum,which is }\\ \left(\frac{1}{1-\frac{1}{3}} \right)\left(\frac{1}{1-\frac{1}{9}} \right)
i hope its right
66@b555 this sum will be 27/8 provided we remove the restriction i≠j≠k
1of course it isnt ;o
but one of the terms of the cube expansion is your expression ;)
1Nopes , just an exercise .
Yes , it does but that ' s not the correct answer as you may very well see .