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∑_(i=1)^nc a_i = ca_1+ ca_2+ ... + ca_n Let's factor out c. ∑_(i=1)^nc a_i = c(a_1+a_2+ ... + a_n) Now, the right hand-side of this equation is c times the sum of the first n terms of the sequence. ∑_(i=1)^nc a_i & = c (a_1+a_2+ ... + a_n_(⇓)) ∑_(i=1)^nc a_i & = c ∑_(i=1)^n a_i ✓ Therefore, the given statement is true.
Commutative Property of Addition
As we can see, the given statement is true.
l Sequence (I): 1,2,3 Sequence (II): 1,2,3 ⇒ l a_n=n b_n=n We will first calculate the sum of the products a_i b_i for i=1, 2, and 3. ∑_(i=1)^3 a_i b_i & = ∑_(i=1)^3 i* i & = 1* 1+ 2* 2+ 3* 3 & = 14 Now we will calculate the product of the series. ∑_(i=1)^3 a_i ∑_(i=1)^3 b_i & = (∑_(i=1)^3 i ) ( ∑_(i=1)^3 i ) & = ( 1+ 2+ 3)( 1+ 2+ 3) & = 36 The calculations give different results. 14 ≠ 36 ⇓ ∑_(i=1)^n a_i b_i ≠ ∑_(i=1)^n a_i ∑_(i=1)^n b_i Therefore, the given statement is false. Note that there are infinitely many counterexamples. We have shown only one of them.
∑_(i=1)^n (a_i)^c ? =( ∑_(i=1)^n a_i )^c
The series on the left-hand side shows the sum of each term raised to the power of c. However, on the right-hand side we have a series raised to the power of c. Intuitively, we may think that they are not equal. Let's consider a finite sequence to check this.∑_(i=1)^3 (a_i)^2 & = ∑_(i=1)^3 i^2 & = 1^2+ 2^2+ 3^2 & = 14 Now we will calculate the square of the sum of the terms. (∑_(i=1)^3 a_i )^2 & = (∑_(i=1)^3 i )^2 & = ( 1+ 2+ 3)^2 & = 36 The calculations give different results. 14 ≠ 36 ⇓ ∑_(i=1)^n (a_i)^c ≠ ( ∑_(i=1)^n a_i )^c Therefore, the given statement is false. There are infinitely many counterexamples to prove this statement is false, and this is just one of them.