C={A1,A2,⋯ ,An},∣Ai∣=i∣⋃i=1nAi∣=∣A1⋃A2⋃⋯⋃An∣=∣A1∣+∣A2∣+⋯+∣An∣Since they are disjoint=1+2+⋯+n=∑i=1ni=n(n+1)2C=\{A_1,A_2,\cdots, A_n\}, |A_i|=i\\ \left| \bigcup_{i=1}^nA_i\right|=\left|A_1 \bigcup A_2 \bigcup \cdots \bigcup A_n\right|\\ =|A_1|+|A_2|+ \cdots + |A_n|\\ \text{Since they are disjoint}\\ =1+2+ \cdots +n\\ = \sum_{i=1}^ni\\ =\frac{n(n+1)}{2}C={A1,A2,⋯,An},∣Ai∣=i∣⋃i=1nAi∣=∣A1⋃A2⋃⋯⋃An∣=∣A1∣+∣A2∣+⋯+∣An∣Since they are disjoint=1+2+⋯+n=∑i=1ni=2n(n+1)
Hence, ∣⋃i=1nAi∣=n(n+1)2.\left| \bigcup_{i=1}^nA_i\right|=\frac{n(n+1)}{2}.∣⋃i=1nAi∣=2n(n+1).
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