Test Index

NCERT Class XI Mathematics - Binomial Theorem - Solutions

© examsnet.com
Question : 7 of 36
Marks: +1, -0
(102)5(102)^5
Solution:  
We have, 102 = 100 + 2
∴ (102)5(102)^5 = (100+2)5(100+2)^5
=  (50)(100)5+ (51)(100)4(2)\,\binom{5}{0}(100)^5+\,\binom{5}{1}(100)^4(2) +  (52)(100)3(2)2+ (53)(100)2(2)3\,\binom{5}{2}(100)^3(2)^2 + \,\binom{5}{3}(100)^2(2)^3 +  (54)(100)(2)4+ (55)(2)5\,\binom{5}{4}(100)(2)^4+\,\binom{5}{5}(2)^5
= (100)5+5(100)4(2)(100)^5+5(100)^4(2) + 10(100)3(4)+10(100)2(8)10(100)^3(4)+10(100)^2(8) + 5 (100) (16) + 1 (32)
= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32
= 11040808032.
© examsnet.com
Go to Question:
Prev Question
Next Question