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NCERT Class XI Mathematics - Binomial Theorem - Solutions

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Question : 3 of 36
Marks: +1, -0
(2x−3)6(2x-3)^6
Solution:  
We have, (2x−3)6(2x-3)^6 = [2x+(−3)]6[2x+(-3)]^6
=  6C0(2x)6+ 6C1(2x)5(−3)\,{}^{6}C_{0}(2x)^6+\,{}^{6}C_{1}(2x)^5(-3) +  6C2(2x)4(−3)2\,{}^{6}C_{2} (2x)^4(-3)^2 +  6C3(2x)3(−3)2\,{}^{6}C_{3}(2x)^3(-3)^2 +  6C4(2x)2(−3)4\,{}^{6}C_{4}(2x)^2(-3)^4 +  6C5(2x)(−3)5\,{}^{6}C_{5}(2x)(-3)^5 +  6C6(−3)6\,{}^{6}C_{6} (-3)^6
= 1(64x6)+6(32x5)(−3)1(64x^6)+6(32x^5)(-3) + 15(16x4)(9)+20(8x3)(−27)15(16x^4)(9)+20(8x^3)(-27) + 15(4x2)(81)15(4x^2)(81) + 6(2x)(−243)+1(729)6(2x)(-243)+1(729)
= 64x6−576x5+2160x4−4320x364x^6-576x^5+2160x^4-4320x^3 + 4860x3+4860x24860x^3+4860x^2 - 2916x + 729
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