The Binomial Question!
Let C 0 , C 1 , C 2 . . . . , denote the binomial coefficients n C 0 , n C 1 , n C 2 . . . so on.
It can be proved that,
( C 1 − C 3 + C 5 − C 7 + ⋯ ) = a b n sin ( c n π ) .
Then find a + b + c .
Details and Assumption
- a , b , c are positive integers.
- a and b have their minimum possible value.
The answer is 8.
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1 solution
Yes ... The same way.....(+1).....
Bonus:-
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It will be similar to the above one. Will just need to substitute ω , ω 2 and 1 in the expansion of ( 1 + x ) n one by one and then add those equations.
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Exactly....
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@Rishabh Jain
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Simple, standard approach. Same way!
The reason I like complex numbers is it gives you two results at the same time, comparing real and imaginary parts.
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@A Former Brilliant Member – True enough! Complex number is quite an interesting,elegant topic. Has wide applications!
2^(n/2) can also be written as 4^(n/4), 8^(n/6) and so on. So answer can be 12, 18 and many more.
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Thanks for that observation and sorry for the mistake. I've edited the question. Please do tell if you still find something wrong.
Thank You
Thanks for the awesome solution!
We know,
( 1 + x ) n = C 0 + C 1 x + C 2 x 2 + . . . + C n x n
Putting x = − ι ( = − 1 ) in the above equation we get,
( 1 − ι ) n = C 0 − C 1 i − C 2 + C 3 i + C 4 + . . . ( − 1 ) n C n i n
Therefore, on writing 1 − ι in its polar form and simplifying we have,
2 2 n ( c o s ( − 4 n π ) + i s i n ( − 4 n π ) ) = ( C 0 − C 2 + C 4 − . . . ) − ι ( C 1 − C 3 + C 5 − . . . )
Now, equating the imaginary parts we get,
C 1 − C 3 + C 5 − . . . = 2 2 n s i n 4 n π
Therefore a + b + c = 8