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NCERT Class XI Mathematics - Binomial Theorem - Solutions

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Question : 14 of 36
Marks: +1, -0
Prove that ∑r=0n3r nCr\sum\limits_{r=0}^{n} 3^r \, {}^nC_r = 4n4^n
Solution:  
We have ,
∑r=0n3r nCr\sum\limits_{r=0}^{n} 3^r \, {}^nC_r = nC30+nC131+nC232{}^nC_3^0 + {}^nC_1 3^1 + {}^nC_2 3^2 + ... + nCn3n{}^nC_n 3^n
= (1+3)n(1+3)^n [Since nC0a0+nC1a{}^nC_0 a^0 + {}^nC_1 a + ... + nCnan{}^nC_n a^n = (1+a)n(1+a)^n] = 4n.
Hence proved.
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