Do you know its property? (4)
If t n = 8 + 6 ( 1 n − 1 ) + 2 ( 2 n − 1 ) , then find the digit sum of n = 1 ∑ 1 7 2 9 ( t n ) .
Details and Assumptions:
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( r n ) = r ! × ( n − r ) ! n ! .
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Digit sum refers to the sum of the digits of the number. For instance, Digit sum of 1729 is 1+7+2+9=19.
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Assume ( r n ) = 0 if n < r .
The answer is 44.
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3 solutions
Did exactly the same way , nice problem @Sandeep Bhardwaj
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thanks ! I created the problem at my own :)
Nice problem bro!
God kill me.... I waste my all chances by due to wrong calculation everytime...!! Anyway Nice Question @Sandeep Bhardwaj as always. But It is My request is that please try to make question with simple calculation... Since I have no calculator. Even You tagged it with JEE so I try to make calculation by my own But fails everytime.
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hehehe..ok. If you are preparing for JEE, its very necessary for you to do as much calculation as you can to get very good RANK. You should not avoid calculations and also should do it correctly with full presence of mind. You recheck you calculation after every step, may be it will help you in reducing your calculation errors. And its a positive point with you that you haven't got a calculator. Keep struggling. All the best!
nice one !! ...... i liked the solution but my procedure was very long!........ im tired :) :p
Dude please please please use \binom{n}{r} in place of ^n C _r
^n C _r appears as n C r
\binom{n}{r} appears as ( r n )
The other symbol that the one you used, is considered standard. I've edited the question, you please edit the solution.
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Ok dude. But what's the issue in using ^nC_r ,as it has been mentioned in the question what does it represent. However, I will edit the solutions. But still I want to know why ?
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Because it makes the wrong sense at times. For example, in your queston, previously it was written
t n = 8 + 6 n − 1 C 1 + 2 n − 1 C 2
Now see, someone (including me) could have thought at first sight that it means
t n = 8 + ( 6 n − 1 ) × C 1 + ( 2 n − 1 ) × C 2
where C 1 and C 2 are some unknown constants that might matter in the question later.
This seems standard, because whenever we are solving (generalising) recurrence relations, such terms (unknown constants multiplied by powers of some numbers) are there.
If you don't believe, just go to Wolfram and see how it shows the output of general term when you input the recurrence
t n = 8 t n − 1 − 1 2 t n − 2
Also, the one you previously used, was not the standard symbol. In no Olympiad or any competition, I've seen use of that one... Hope I've convinced why \binom{n}{r} should be used.
If you want it to appear as a bigger and clearer one, use \dbinom{n}{r}
Reply please , friend :)
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@Aditya Raut – Ok friend. Thank you very much, I got what you wanted to convey. I will apply the things you told me and will try to improve my latex coding. :)
One more thing @Sandeep Bhardwaj , there are some fundamental things about summation you gotta see.
Never use the same variable as the range, like
n = 1 ∑ n ( 2 n + 1 )
Please use it as k = 1 ∑ n ( 2 k + 1 )
This is because always, the n will be a number which will be the upper limit of the sum. And you can't use it as a variable. Use another variable, and because your upper limit will be n , the v a r i a b l e will become into that value in the final answer.
Also please use \displaystyle before the \sum symbol, (see the changes that I've made to your solution)
When you write
\sum {n=0}^{1729} t n , it appears as ∑ n = 0 1 7 2 9 t n
\displaystyle at start makes it appears as n = 0 ∑ 1 7 2 9 t n
I think it missed something. Look at the last assumption. By doing it your way, you done it wrong since you've neglected some terms. It should then be 1731890886 or 51 as the digit sum.
I bashed t n = n 2 + 3 n + 4 , and found the summation manually :P.
i got 1728903166 , god knows why i got wrong , also is it necessary to shove 1729 in every problem ? ...sigh... ramanujan fans :P
We can observe by expansion of the terms that
n = 1 ∑ 1 7 2 9 ( 8 + 6 ( 1 n − 1 ) + 2 ( 2 n − 1 ) ) = n = 1 ∑ 1 7 2 9 ( n 2 + 3 n + 4 )
= n = 1 ∑ 1 7 2 9 ( n 2 ) + 3 n = 1 ∑ 1 7 2 9 ( n ) + n = 1 ∑ 1 7 2 9 ( 4 )
= 1 7 2 4 4 0 9 5 0 5 + 3 × 1 4 9 5 5 8 5 + 4 × 1 7 2 9
= 1 7 2 4 4 0 9 5 0 5 + 4 4 8 6 7 5 5 + 6 9 1 6
= 1 7 2 8 9 0 3 1 7 6
1 + 7 + 2 + 8 + 9 + 0 + 3 + 1 + 7 + 6 = 4 4
Got the result, but did not use this property of binomial coefs. The first binomial coef equals to n-1, the second -- (n-1)(n-2)/2. The expression in the sum is n^2 +3n +4.
The sums for each term are:
Sum(n^2) = n(n+1)(2n+1)/6 = 1729(1729+1)(2 * 1729+1)/6.
3*Sum(n) = 3 * n(n+1)/2=3 * 1729(1729+1) / 2.
Sum(4) = 4 * 1729.
The total sum is 1728903176. The sum of the digits 44.
From the properties of Binomial coefficient, we have k = 1 ∑ n ( k n + k − 1 ) = ( k + 1 n + k ) for any given natural number k
For example, k = 1 ∑ n ( 2 n + 1 ) = ( 3 n + 2 )
So here k = 1 ∑ n 8 + 6 ( 1 n − 1 ) + 2 ( 2 n − 1 ) = 8 n + 6 ( 2 n ) + 2 ( 3 n ) using the above stated property
Putting n = 1 7 2 9 , k = 1 ∑ n 8 + 6 ( 1 n − 1 ) + 2 ( 2 n − 1 ) = 1 7 2 8 9 0 3 1 7 6
So digit sum here is 1 + 7 + 2 + 8 + 9 + 0 + 3 + 1 + 7 + 6 = 4 4
enjoy !