Question

Question 12 Find (x + 1)^6 + (x – 1)^6. Hence or otherwise evaluate [ 2^(1/2) + 1 ]^6 + [ 2^(1/2) - 1 ]^6.

Class X1 - Maths -Binomial Theorem Page 167

Answer 1

(x +1)^6
Use the formula,
(x + y)^n = nC0.x^n + nC1.x^(n-1)y + nC2.x^(n-2)y^2 + ........+nCn.y^n

(x+1)^6 = 6C0.x^6 + 6C1.x^5 + 6C2.x^4 + 6C3.x^3 + 6C4.x^2 + 6C5.x + 6C6.(1)^6
= x^6 + 6x^5 + 15x^4 +20x^3 + 15x^2 + 6x + 1------(1)

Similarly ,
(x - 1)^6 = x^6 - 6x^5 + 15x^4 -20x^3 + 15x^2 - 6x + 1----------(2)

Now add both equations,

(x+1)^6 +(x-1)^6 = (x^6+6x^5+15x^4+20x^3+15x^2+6x+1)+(x^6-6x^5+15x^4-20x^3+15x^2-6x+1)
= 2[x^6+15x^4+15x^2+1]

Hence,
(√2+1)^6 + (√2+1)^6 =2[√2^6 + 15√2^4 + 15√2^2 + 1]
= 2[8 + 60 + 30 + 1]
= 2[ 99]
= 198