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Using binomial theorem find (i) (101)^(5) (ii) 51^(6) |
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Answer» Solution :(i) We can WRITE `(101)^(5)=(100+1)^(5)` Now by using binomial theorem, we GET `(100+1)^(5)=(100)^(5)+.^(5)C_(1)(100)^(4)+.^(5)C_(2)100^(3)+.^(5)C_(3)100^(2)+.^(5)C_(4)(100)+.^(5)C_(5)` `=10000000000+5xx100000000+10xx1000000+10xx10000+5xx100+1` `=10000000000+500000000+10000000+100000+500+1` `=1051010001` (ii) We can write `(51)^(6)=(50+1)^(6)` By using binomial theorem, we get `(1+50)^(6)=1+.^(6)C_(1)50+.^(6)C_(2)(50)^(2)+.^(6)C_(3)(50)^(3)+.^(6)C_(4)(50)^(4)+.^(6)C_(5)(50)^(5)+.^(6)C_(6)(50)^(6)` `=1+6xx50+15xx2500+20xx125000+15xx6250000+6xx312500000+15625000000` `=1+300+37500+2500000+93750000+1875000000+15625000000` `=17596287801` |
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