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NCERT Class XI Mathematics - Binomial Theorem - Solutions

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Question : 12 of 36
Marks: +1, -0
Find (x+1)6+(x−1)6(x+1)^6+(x-1)^6. Hence or otherwise evaluate (2+1)6+(2−1)6(\sqrt{2}+1)^6+(\sqrt{2}-1)^6.
Solution:  
By using binomial theorem, we have
(x+1)6+(x−1)6(x+1)^6+(x-1)^6 = [6C0x6+6C1x5+6C2x4{}^{6}C_{0}x^6+{}^{6}C_{1}x^5+{}^{6}C_{2}x^4 +   ⁣6C3x3+  ⁣6C4x2+  ⁣6C5x+  ⁣6C6\,\!{}^{6}C_{3}x^3+\,\!{}^{6}C_{4}x^2+\,\!{}^{6}C_{5}x+\,\!{}^{6}C_{6}] + [6C0+x6−6C1x5+6C2x4{}^6C_{0}+x^6-{}^6C_{1}x^5+{}^6C_{2}x^4 - 6C3x3+6C4x2−6C5x+6C6{}^{6}C_{3}x^3+{}^{6}C_{4}x^2-{}^{6}C_{5}x+{}^{6}C_{6}]
= x6+6x5+15x4+20x3+15x2x^6+6x^5+15x^4+20x^3+15x^2 + 6x + 1 + x6−6x5+15x4−20x3+15x2x^6-6x^5+15x^4-20x^3+15x^2 - 6x + 1
∴ (x+1)6+(x−1)6(x+1)^6+(x-1)^6 = 2x6+30x4+30x22x^6+30x^4+30x^2 + 2 ... (i)
Substituting x = 2\sqrt{2}, in (i), we get
(2+1)6+(2−1)6(\sqrt{2}+1)^6+(\sqrt{2}-1)^6 = 2(2)6+30(2)4+30(2)2+22(\sqrt{2})^6+30(\sqrt{2})^4+30(\sqrt{2})^2+2
= 16 + 120 + 60 + 2 = 198
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