Question

Question 7 Using Binomial Theorem, evaluate (102)^5

Class X1 - Maths -Binomial Theorem Page 167

Answer 1

(102)^5
use the formula,
(x + y)^n = nC0.x^n + nC1.x^(n-1)y + nC2.x^(n-2)y^2 + ........+nCn.y^n


(102)^5 = (100 + 2)^5
(100 + 2)^5 = 5C0.(100)^5 + 5C1.(100)^4.(2) + 5C2.(100)^3.(2)^2 + 5C3.(100)^2.(2)^3 + 5C4.(100)(2)^4 + 5C5(2)^5

= 5C0.(10)^10 + 5C1.(10^8)(2) + 5C2.(10^6) × 4 + 5C3.(10^4) × 8 + 5C4.(10^2) × 16 + 5C5 × 32

= {5!/5!} 10^10 + {5/4!} × 2 × 10^8 + {5!/2!×3!} × 4 × 10^6 + {5!/3!×2!} × 8 × 10^4 + {5!/4!} × 16 × 10^2 + {5!/5!} × 32

= 10^10 + 5 × 10^8 × 2 + {5×4/2} × 10^6 × 4 + {5×4/2} × 10^4 × 8 + 5 × 100 × 16 + 32

= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32

= 11040808032

Answer 2

Answer:

answer is 11040808032