1+2+3+... +65 = 65*66/2 = 65*33 = 2145
For the required sum, each term is simply double the corresponding term above, so
2+4+6+... + 130 = 2*2145 = 4290
2048
78 + 50 = 128
The answer is 545
545
x = 64 64 + 64 = 128
what if someone has a number already to add with another number to equal 128, so 64+64 still equals 128.
Equilibrium occurs where Qd = QsQd - 128 + 9p = 0Qd = 128 - 9pQs + 32 - 7p = 0Qs = 7p - 327p - 32 = 128 - 9p16p = 160p = 10
Well, I would like to tell you that it is not 48 but it is 128
67 times 2 and 134 times 1
128
The answer to that is 256.
There is no r in 128 plus 253