NCERT Class XI Mathematics - Binomial Theorem - Solutions | Question 33

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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