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∑10Sp=(40)2p=1∑10p−(10)(20)(39)=(40)22(10)(11)−(10)(20)(39)=80,200.