217k views
0 votes
Hello.
How to prove that those operations are truth?
Sums

Hello. How to prove that those operations are truth? Sums-example-1

1 Answer

5 votes

a. This is just shifting the index of summation:


\displaystyle \sum_(j=0)^4 d_(j+1) = d_(0+1)+d_(1+1)+\cdots+d_(4+1) \\ = d_(2-1)+d_(3-1)+\cdots+d_(6-1) \\ = \sum_(j=2)^6 d_(j-1)

and just swap j with k, since you can freely use any symbol for the index.

That is, the identity relies on simple integer arithmetic: 0 + 1 = 1 = 2 - 1, and 1 + 1 = 2 = 3 - 1, and so on.

b. This is something of a generalization of the identity in part a.


\displaystyle \sum_(j=1)^n a_jb_(m+j) = a_1b_(m+1) + a_2b_(m+2) + \cdots + a_nb_(m+n) \\ = a_((m+1)-m)b_(m+1) + a_((m+2)-m)b_(m+2) + \cdots + a_((m+n)-m)b_(m+n) \\ = \sum_(j=m+1)^(m+n) a_(j-m)b_j

c. Distribute the summation to get


\displaystyle \sum_(j=1)^m (6d_j+1) = \sum_(j=1)^m 6d_j + \sum_(j=1)^m 1

The second sum here is just adding together m copies of 1,


\displaystyle \sum_(j=1)^m 1 = \underbrace{1+1+\cdots+1}_(m\,\rm times) = m

so


\displaystyle \sum_(j=1)^m (6d_j+1) = \sum_(j=1)^m 6d_j + m

d. The sum as given is generally not true:


\displaystyle \sum_(n=1)^k a_k = a_1+a_2+\cdots+a_k

while (generalizing the identity from part c)


\displaystyle \sum_(m=2)^(k+1) (a_m-1) = (a_2+a_3+\cdots+a_k+a_(k+1))-k

and these expressions are equal only if
a_1=a_(k+1)-k.

User Geroge
by
8.4k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories