Problem: For p=1,2,…,10 let Sp​ be the sum of the first 40 terms of the arithmetic progression whose first term is p and whose common difference is 2p−1; then S1​+S2​+…+S10​ is
Answer Choices:
A. 80,000
B. 80,200
C. 80,400
D. 80,600
E. 80,800
Solution:
For each p=1,⋯,10,
Sp​​=p+p+(2p−1)+p+2(2p−1)+⋯+p+39(2p−1)​=40p+2(40)(39)​(2p−1)​=(40+40⋅39)p−20⋅39​=(40)2p−20⋅39​
​ Therefore, ​p=1∑10​Sp​​=(40)2p=1∑10​p−(10)(20)(39)​=(40)22(10)(11)​−(10)(20)(39)​=80,200.​​