Answer:
11∗2+13∗4+15∗6+⋯+199∗100=?
=(1−12)+(13−14)+(15−16)+⋯+(199−1100)
=(1+13+15+17+⋯+199)−(12+14+16+⋯+1100)
=(1+13+15+17+⋯+199)−12(1+12+13+⋯+150)
=∑k=15012k−1−∑k=15012k
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