


Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Last updated at May 29, 2023 by Teachoo
Ex 7.1, 14 (Method 1) By Binomial Theorem, Putting b = 3 and a = 1 in the above equation Prove that β_(π=0)^πβγ3^π nCrγ β_(π=0)^πβnCr π^(π β π) π^π β_(π=0)^πβnCr 1^(πβπ) 3^π Hence proved Ex 7.1, 14 (Method 2) β Introduction For r = 0, 3^0 nC0 For r = 1, 3^1 nC1 For r = 2, 3^2 nC2 For r = 3, 3^3 nC3 β¦ β¦. For r = n, 3^π nCn nC0 30 + nC1 31 + nC2 32 + β¦ β¦β¦β¦ + nCn β 1 3n β 1 + nCn 3n Prove that = nC0 30 + nC1 31 + nC2 32 + β¦β¦β¦β¦β¦β¦ + nCn-1 3n-1 + nCn 3n Ex 7.1, 14(Method 2) Solving L.H.S This is similar to nC0 an b0 + nC1 an-1 b1 + nC2 an-2 b2 + β¦β¦ .+ nCn-1 a1 bn-1 + nCn a0 bn Where a = 1 , b = 3 And we know that (a + b)n = nC0 an b0 + nC1 an-1 b1 + β¦β¦.+ nCn-1 a1 bn-1 + nCn a0 bn = (1 + 3)n = (4)n = R.H.S Hence proved