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NCERT Class XI Mathematics - Binomial Theorem - Solutions

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Question : 33
Total: 36
Find an approximation of (0.99)5 using the first three terms of its expansion.
Solution:  
We have, 0.99 = (1 – 0.01)
∴ (0.99)5 = (1−0.001)5
=
5
‌
C0+
5
‌
C1(−0.01)+
5
‌
C2(−0.01)2+
5
‌
C3(−0.01)3+
5
‌
C4(−0.01)4+
5
‌
C5(−0.01)5

=1+5(−0.01)+10(0.0001)+
5
‌
C3(−0.01)3+
5
‌
C4(−0.01)4+
5
‌
C5(−0.01)5

=0.951
5
‌
C3(−0.01)3+
5
‌
C4(−0.01)4+
5
‌
C5×(−0.01)5

Hence, (0.99)5 is nearly equal to 0.951.
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