How is the summation being expanded? I am trying to understand summations by solving some example p

cooloicons62 2022-07-07 Answered
How is the summation being expanded?
I am trying to understand summations by solving some example problems, but I could not understand how is the second to last line being expanded? I would really appreciate if you could explain me how is it being expanded.
i = 1 n 1 j = i + 1 n k = 1 j 1 = i = 1 n 1 j = i + 1 n j = i = 1 n 1 ( j = 1 n j j = 1 i j ) = i = 1 n 1 ( n ( n + 1 ) 2 i ( i + 1 ) 2 ) = 1 2 i = 1 n 1 n 2 + n i 2 i = 1 2 ( ( n 1 ) n 2 + ( n 1 ) n ( n ( n + 1 ) ( 2 n + 1 ) 6 n 2 ) ( n ( n + 1 ) 2 n ) ) = f ( n ) = n ( n ( n + 1 ) ) 2 n ( n + 1 ) ( 2 n + 1 ) 12 n ( n + 1 ) 4
You can still ask an expert for help

Want to know more about Discrete math?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

Keegan Barry
Answered 2022-07-08 Author has 18 answers
Step 1
I take it that what has to be explained is this (I've introduced parentheses on the left hand side for clarity):
i = 1 n 1 ( n 2 + n i 2 i ) = ( n 1 ) n 2 + ( n 1 ) n ( n ( n + 1 ) ( 2 n + 1 ) 6 n 2 ) ( n ( n + 1 ) 2 n ) .
This equation results from adding together the following four identities:
i = 1 n 1 n 2 = ( n 1 ) n 2 , i = 1 n 1 n = ( n 1 ) n , i = 1 n 1 i 2 = i = 1 n i 2 n 2 = n ( n + 1 ) ( 2 n + 1 ) 6 n 2 , i = 1 n 1 i = i = 1 n i n = n ( n + 1 ) 2 n .
Step 1
Lines 4 and 6 follow, of course, from the familiar identities:
i = 1 n i 2 = n ( n + 1 ) ( 2 n + 1 ) 6 , i = 1 n i = n ( n + 1 ) 2 .
I don't know why it was done this way! It seems to me that it would have been simpler just to write:
i = 1 n 1 i 2 = ( n 1 ) n ( 2 n 1 ) 6 , i = 1 n 1 i = ( n 1 ) n 2 .
(Also, in the comments, I've suggested two ways to arrive at the final answer with less calculation.)

We have step-by-step solutions for your answer!

kolutastmr
Answered 2022-07-09 Author has 2 answers
Step 1
Starting from line 4 of the displayed equation, we have
i = 1 n 1 n ( n + 1 ) 2 = ( n 1 ) n ( n + 1 ) 2 because there are n 1 equal terms. Also, i = 1 n 1 i ( i + 1 ) 2 = i = 1 n 1 ( i ( i + 1 ) ( i + 2 ) 6 ( i 1 ) i ( i + 1 ) 6 ) is a telescoping sum, hence equal to ( n 1 ) n ( n + 1 ) 0 6 .
Step 2
Therefore, the whole sum is equal to ( n 1 ) n ( n + 1 ) 2 ( n 1 ) n ( n + 1 ) 6 = ( n 1 ) n ( n + 1 ) 3 .
The "weird" decomposition above comes in fact from a general property of the raising factorials: e.g.
i ( i + 1 ) ( i + 2 ) ( i + 3 ) ( i + 4 ) ( i 1 ) i ( i + 1 ) ( i + 2 ) ( i + 3 ) = 5 i ( i + 1 ) ( i + 2 ) ( i + 3 ) .

We have step-by-step solutions for your answer!

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2021-07-28

Let A, B, and C be sets. Show that (AB)C=(AC)(BC)
image

asked 2021-08-15
How many elements are in the set { 0, { { 0 } }?
asked 2021-08-18
Discrete Mathematics Basics
1) Determine whether the relation R on the set of all Web pages is reflexive, symmetric, antisymmetric, and/or transitive, where (a,b)R if and only if
I) everyone who has visited Web page a has also visited Web page b.
II) there are no common links found on both Web page a and Web page b.
III) there is at least one common link on Web page a and Web page b.
asked 2020-11-09
Use proof by Contradiction to prove that the sum of an irrational number and a rational number is irrational.
asked 2022-09-07
Discrete Math Help with a Proof
I need help to prove the following: Let a, b, and c be any integers. If a∣b, then a∣bc
asked 2022-06-26
Number of combinations with given outcomes for a set number of dice rolls
How to find the number of combinations for n rolls of a pair of 6-sided dice with all of the specified sums rolled at least once?
Example. A pair of 6-sided dice is rolled 5 times. What is the number of combinations containing each of: 3, 4, 8 at least once? So that (3, 7, 8, 5, 4) is counted and (3, 4, 2, 2, 12) is not.
Thoughts on the example. It's possible to roll 3 in 2 ways, 4 in 3 and 8 in 5 and the total number of outcomes is 36. So I think the number of combinations containing 3, 4, 8 is:
C = 36 2 × 5 P 3 × 2 × 3 × 5 D
Where D is duplicates. But if this is correct, not sure what pattern the duplicates follow.
asked 2022-07-07
How many 20-digit numbers can be formed from { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } such that no 2 consecutive digit is both odd
I've noticed that the number of odd digit in the number must be less than 11. But i can't progress more than that.

New questions

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question