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NCERT Class XI Mathematics - Binomial Theorem - Solutions

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Question : 33 of 36
Marks: +1, -0
Find an approximation of (0.99)5(0.99)^5 using the first three terms of its expansion.
Solution:  
We have, 0.99 = (1 – 0.01)
∴ (0.99)5(0.99)^5 = (1−0.001)5(1-0.001)^5
=
 5C0+ 5C1(−0.01)+ 5C2(−0.01)2+ 5C3(−0.01)3+ 5C4(−0.01)4+ 5C5(−0.01)5\,{}^{5}C_0+\,{}^{5}C_1(-0.01)+\,{}^{5}C_2(-0.01)^2 + \,{}^{5}C_3(-0.01)^3+\,{}^{5}C_4(-0.01)^4+\,{}^{5}C_5(-0.01)^5
=1+5(−0.01)+10(0.0001)+ 5C3(−0.01)3+ 5C4(−0.01)4+ 5C5(−0.01)5= 1 + 5 (-0.01) + 10 (0.0001) + \,{}^{5}C_3 (-0.01)^3 + \,{}^{5}C_4(-0.01)^4+\,{}^{5}C_5(-0.01)^5
=0.951 5C3(−0.01)3+5C4(−0.01)4+5C5×(−0.01)5= 0.951\,{}^{5}C_3 (-0.01)^3+{}^{5}C_4(-0.01)^4+{}^{5}C_5 \times (-0.01)^5
Hence, (0.99)5(0.99)^5 is nearly equal to 0.951.
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