{"id":231160,"date":"2025-06-10T14:33:27","date_gmt":"2025-06-10T14:33:27","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=231160"},"modified":"2025-06-10T14:33:29","modified_gmt":"2025-06-10T14:33:29","slug":"the-hockeystick-identity-is-given-by","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/10\/the-hockeystick-identity-is-given-by\/","title":{"rendered":"The hockeystick identity is given by"},"content":{"rendered":"\n

The hockeystick identity is given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Here is the solution to the problem.<\/p>\n\n\n\n

(a) Proof using Pascal’s identity<\/h3>\n\n\n\n

The hockey-stick identity is <\/p>\n\n\n\n

\u2211k=0r(n+kk)=(n+r+1r)\u2211k<\/em>=0r<\/em>\u200b(k<\/em>n<\/em>+k<\/em>\u200b)=(r<\/em>n<\/em>+r<\/em>+1\u200b)<\/code><\/pre>\n\n\n\n
. This proof proceeds by repeatedly applying Pascal’s identity, which states <\/p>\n\n\n\n
(mj)+(mj\u22121)=(m+1j)(j<\/em>m<\/em>\u200b)+(j<\/em>\u22121m<\/em>\u200b)=(j<\/em>m<\/em>+1\u200b)<\/code><\/pre>\n\n\n\n
.<\/p>\n\n\n\n
The proof begins with the left-hand side (LHS) of the identity. The first term in the summation, <\/p>\n\n\n\n
(n0)(0n<\/em>\u200b)<\/code><\/pre>\n\n\n\n
, is equal to 1. Since <\/p>\n\n\n\n
nn<\/em><\/code><\/pre>\n\n\n\n
 is a positive integer, <\/p>\n\n\n\n
(n+10)(0n<\/em>+1\u200b)<\/code><\/pre>\n\n\n\n
 is also equal to 1. Thus, <\/p>\n\n\n\n
(n0)(0n<\/em>\u200b)<\/code><\/pre>\n\n\n\n
 can be replaced by <\/p>\n\n\n\n
(n+10)(0n<\/em>+1\u200b)<\/code><\/pre>\n\n\n\n
 without changing the value of the sum.<\/p>\n\n\n\n
The LHS becomes:<\/p>\n\n\n\n
(n+10)+(n+11)+(n+22)+\u22ef+(n+rr)(0n<\/em>+1\u200b)+(1n<\/em>+1\u200b)+(2n<\/em>+2\u200b)+\u22ef+(r<\/em>n<\/em>+r<\/em>\u200b)<\/code><\/pre>\n\n\n\n
Applying Pascal’s identity to the first two terms gives <\/p>\n\n\n\n
(n+10)+(n+11)=(n+21)(0n<\/em>+1\u200b)+(1n<\/em>+1\u200b)=(1n<\/em>+2\u200b)<\/code><\/pre>\n\n\n\n
. The sum is now:<\/p>\n\n\n\n
(n+21)+(n+22)+(n+33)+\u22ef+(n+rr)(1n<\/em>+2\u200b)+(2n<\/em>+2\u200b)+(3n<\/em>+3\u200b)+\u22ef+(r<\/em>n<\/em>+r<\/em>\u200b)<\/code><\/pre>\n\n\n\n
Again, applying Pascal’s identity to the new first two terms gives <\/p>\n\n\n\n
(n+21)+(n+22)=(n+32)(1n<\/em>+2\u200b)+(2n<\/em>+2\u200b)=(2n<\/em>+3\u200b)<\/code><\/pre>\n\n\n\n
. This process, known as a telescoping sum, continues. Each step combines the first two terms into a single term. After <\/p>\n\n\n\n
r\u22121r<\/em>\u22121<\/code><\/pre>\n\n\n\n
 steps, the sum is reduced to two terms:<\/p>\n\n\n\n
(n+rr\u22121)+(n+rr)(r<\/em>\u22121n<\/em>+r<\/em>\u200b)+(r<\/em>n<\/em>+r<\/em>\u200b)<\/code><\/pre>\n\n\n\n
A final application of Pascal’s identity yields <\/p>\n\n\n\n
(n+r+1r)(r<\/em>n<\/em>+r<\/em>+1\u200b)<\/code><\/pre>\n\n\n\n
, which is the right-hand side of the identity.<\/p>\n\n\n\n
(b) Proof by combinatorial argument<\/h3>\n\n\n\n
The identity is first rewritten using the symmetry property <\/p>\n\n\n\n
(NK)=(NN\u2212K)(K<\/em>N<\/em>\u200b)=(N<\/em>\u2212K<\/em>N<\/em>\u200b)<\/code><\/pre>\n\n\n\n
. The original identity <\/p>\n\n\n\n
\u2211k=0r(n+kk)=(n+r+1r)\u2211k<\/em>=0r<\/em>\u200b(k<\/em>n<\/em>+k<\/em>\u200b)=(r<\/em>n<\/em>+r<\/em>+1\u200b)<\/code><\/pre>\n\n\n\n
 becomes:<\/p>\n\n\n\n
\u2211k=0r(n+kn)=(n+r+1n+1)k<\/em>=0\u2211r<\/em>\u200b(n<\/em>n<\/em>+k<\/em>\u200b)=(n<\/em>+1n<\/em>+r<\/em>+1\u200b)<\/code><\/pre>\n\n\n\n
This form can be proven by a combinatorial argument. Consider the problem of counting the number of ways to choose a subset of <\/p>\n\n\n\n
n+1n<\/em>+1<\/code><\/pre>\n\n\n\n
 objects from a set of <\/p>\n\n\n\n
n+r+1n<\/em>+r<\/em>+1<\/code><\/pre>\n\n\n\n
 distinct objects arranged in a line.<\/p>\n\n\n\n
The right-hand side (RHS), <\/p>\n\n\n\n
(n+r+1n+1)(n<\/em>+1n<\/em>+r<\/em>+1\u200b)<\/code><\/pre>\n\n\n\n
, counts this number directly, by the definition of a combination.<\/p>\n\n\n\n
The left-hand side (LHS) counts the same quantity by partitioning the selections into <\/p>\n\n\n\n
r+1r<\/em>+1<\/code><\/pre>\n\n\n\n
 disjoint cases. The partition is based on the position of the leftmost object chosen from the line. Let the position of this object be <\/p>\n\n\n\n
pp<\/em><\/code><\/pre>\n\n\n\n
. The remaining <\/p>\n\n\n\n
nn<\/em><\/code><\/pre>\n\n\n\n
 objects must then be selected from the <\/p>\n\n\n\n
(n+r+1)\u2212p(n<\/em>+r<\/em>+1)\u2212p<\/em><\/code><\/pre>\n\n\n\n
 objects to its right.<\/p>\n\n\n\n
For a selection to be possible, there must be at least <\/p>\n\n\n\n
nn<\/em><\/code><\/pre>\n\n\n\n
 objects to the right of position <\/p>\n\n\n\n
pp<\/em><\/code><\/pre>\n\n\n\n
. This implies <\/p>\n\n\n\n
(n+r+1)\u2212p\u2265n(n<\/em>+r<\/em>+1)\u2212p<\/em>\u2265n<\/em><\/code><\/pre>\n\n\n\n
, which simplifies to <\/p>\n\n\n\n
p\u2264r+1p<\/em>\u2264r<\/em>+1<\/code><\/pre>\n\n\n\n
. The position <\/p>\n\n\n\n
pp<\/em><\/code><\/pre>\n\n\n\n
 can therefore range from <\/p>\n\n\n\n
11<\/code><\/pre>\n\n\n\n
 to <\/p>\n\n\n\n
r+1r<\/em>+1<\/code><\/pre>\n\n\n\n
.<\/p>\n\n\n\n
For each case where the leftmost chosen object is at position <\/p>\n\n\n\n
pp<\/em><\/code><\/pre>\n\n\n\n
, there are <\/p>\n\n\n\n
(n+r+1\u2212pn)(n<\/em>n<\/em>+r<\/em>+1\u2212p<\/em>\u200b)<\/code><\/pre>\n\n\n\n
 ways to choose the other <\/p>\n\n\n\n
nn<\/em><\/code><\/pre>\n\n\n\n
 objects. Summing over all possible values of <\/p>\n\n\n\n
pp<\/em><\/code><\/pre>\n\n\n\n
 gives the total number of ways:<\/p>\n\n\n\n
\u2211p=1r+1(n+r+1\u2212pn)p<\/em>=1\u2211r<\/em>+1\u200b(n<\/em>n<\/em>+r<\/em>+1\u2212p<\/em>\u200b)<\/code><\/pre>\n\n\n\n
By performing a change of index with <\/p>\n\n\n\n
k=r+1\u2212pk<\/em>=r<\/em>+1\u2212p<\/em><\/code><\/pre>\n\n\n\n
, as <\/p>\n\n\n\n
pp<\/em><\/code><\/pre>\n\n\n\n
 ranges from <\/p>\n\n\n\n
11<\/code><\/pre>\n\n\n\n
 to <\/p>\n\n\n\n
r+1r<\/em>+1<\/code><\/pre>\n\n\n\n
, the index <\/p>\n\n\n\n
kk<\/em><\/code><\/pre>\n\n\n\n
 ranges from <\/p>\n\n\n\n
rr<\/em><\/code><\/pre>\n\n\n\n
 down to <\/p>\n\n\n\n
00<\/code><\/pre>\n\n\n\n
. The summation becomes <\/p>\n\n\n\n
\u2211k=0r(n+kn)\u2211k<\/em>=0r<\/em>\u200b(n<\/em>n<\/em>+k<\/em>\u200b)<\/code><\/pre>\n\n\n\n
, which is the LHS. Since both sides count the same set of combinations, the identity is proven.<\/p>\n\n\n\n
\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
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