1.8.6 Example F Use Theorem 1.6 to evaluate , k· 2k. Comparison of this expression with equation (1.249) gives A = 2 and Pki k. Since AP = 1 and A" Pk k=1 O for m > 1, we have from equation (1.249) 2k...

Explain and the theorem 1.6 this

1, we have from equation (1.249) 2k 2.1 A-12kk = E2*k : + 0+ +0+ 2 – 1 2 – 1 (1.273) - = 2* (k – 2). Applying the fundamental theorem of the sum calculus gives n Ek2* = A-1 (k2*) +1 k=1 (1.274) = 2* (k – 2)|+ = 2n+1 (n – 1) + 2. "/>
Extracted text: 1.8.6 Example F Use Theorem 1.6 to evaluate , k· 2k. Comparison of this expression with equation (1.249) gives A = 2 and Pki k. Since AP = 1 and A" Pk k=1 O for m > 1, we have from equation (1.249) 2k 2.1 A-12kk = E2*k : + 0+ +0+ 2 – 1 2 – 1 (1.273) - = 2* (k – 2). Applying the fundamental theorem of the sum calculus gives n Ek2* = A-1 (k2*) +1 k=1 (1.274) = 2* (k – 2)|+ = 2n+1 (n – 1) + 2.


Extracted text: Theorem 1.6. For A# 1, (1.249) (A – 1)2