Solveeit Logo
Login

Question

Mathematics Question on Binomial Theorem for Positive Integral Indices

Find (x+1)6−(x−1)6(x + 1)^6 - (x - 1)^6. Hence or otherwise evaluate (2+1)6+(2−1)6.(\sqrt2 + 1)^6 + (\sqrt2 - 1)^6.

Answer

Using Binomial Theorem, the expressions, (x+1)6(x+ 1)^6 and (x−1)6(x -1)^6, can be expanded as

(x+1)6(x+1)^6= ^6C_0 x^6+ ^6C_1 x^5$$+ ^6C_2 x^4 + ^6C_3 x^3$$+ ^6C_4 x^2 + ^6C_5 x + ^6C_6
(x−1)6(x-1)^6= ^6C_0 x^6 - ^6C_1 x^5 + ^6C_2 x^4$$- ^6C_3 x^3$$+ ^6C_4 x^2 - ^6C_5 x + ^6C_6
(x+1)6+(x−1)6(x+1)^6+(x-1)^6=2[6C0x62[ ^6C_0 x^6 +6C2x4++ ^6C_2 x^4 + 6C4x2+^6C_4 x^2+ 6C6]^6C_6]
=2[x6+15x4+15x2+1]2[x^6+15x^4+15x^2+1]

By putting x=2,we obtain\text{By putting } x = \sqrt2, \text{we obtain}

(2+1)6+(2−1)6(\sqrt2+1)^6+(\sqrt2-1)^6 = 2[(2)6+15(2)4+15(2)2+1]2[ (\sqrt2)^6 +15(\sqrt2)^4 +15 (\sqrt2)^2+1]
=2(8+15×4+15×2+1)2(8+15×4+15 ×2+1)
=2(8+60+30+1)2(8+60+30+1)
=2(99)2(99)
=198198

Hence, (2+1)6+(2−1)6=198.\text {Hence,} \,(\sqrt2 + 1)^6 + (\sqrt2 - 1)^6 = 198.