n = 0 ∑ ∞ n + 1 1 k = 0 ∑ n ( n k ) k + 1 ( − 1 ) k = b π a
The equation above holds true for integers a and b . Submit your answer as a + b .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
@Naren Bhandari I'm not sure about the argument with the telescopic sum at the end - shouldn't it be k ≤ 2 n from the second sum onwards? Also the binomial coefficients don't telescope because one coefficient is odd and the other is even...
It might be easier not to split the sum into even and odd parity and just use the "Binomial Theorem": k = 0 ∑ n ( k n ) k + 1 ( − 1 ) k = n + 1 1 k = 0 ∑ n ( k + 1 n + 1 ) ( − 1 ) k = n + 1 1 ( 1 − k = 0 ∑ n + 1 ( k n + 1 ) ( − 1 ) k )
Log in to reply
Thank you for spotting it out. I didnt check last steps of mine throughly since so I made comment of telescoping( which is a mistake I did). I have updated solution now.
I just saw the form b π a which reminded me of the Basel problem so I guessed 6 π 2
Problem Loading...
Note Loading...
Set Loading...
Since k = 0 ∑ n ( k n ) k + 1 ( − 1 ) k = ∫ 0 1 ( k = 0 ∑ n ( k n ) ( − x ) k ) d x = ∫ 0 1 ( 1 − x ) n d x = ∫ 0 1 x n d x = n + 1 1 which give the final closed form n ≥ 0 ∑ ( n + 1 ) 2 1 = n ≥ 1 ∑ n 2 1 = 6 π 2
The finite sum can be observed as follows by splitting the sum in even and odd parity f k ≥ 0 ( n ) = k ≤ n ∑ ( k n ) k + 1 ( − 1 ) k = k ≤ n ∑ ( ( 2 k n ) 2 k + 1 1 − ( 2 k + 1 n ) 2 k + 2 1 ) = k ≤ n ∑ ( ( 2 k + 1 n + 1 ) − ( 2 k + 2 n + 1 ) ) n + 1 1 = n + 1 1 Alternative way to go is explained by Carsten Meyer in comment section.