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JEE Main 2018
Binomial Theorem
Binomial Theorem
Hard

Question

Let s1=j=110j(j1)10Cj{s_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right){}^{10}} {C_j}, s2=j=110j.10Cj{{s_2} = \sum\limits_{j = 1}^{10} {} } j.{}^{10}{C_j} and s3=j=110j2.10Cj.{{s_3} = \sum\limits_{j = 1}^{10} {{j^2}.{}^{10}{C_j}.} } Statement-1 : S3=55×29{{S_3} = 55 \times {2^9}}. Statement-2 : S1=90×28{{S_1} = 90 \times {2^8}} and S2=10×28{{S_2} = 10 \times {2^8}}.

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Solution

1. Introduction to Key Binomial Sum Identities

This problem involves evaluating sums of the form jnCj\sum j \cdot {}^n C_j, j(j1)nCj\sum j(j-1) \cdot {}^n C_j, and j2nCj\sum j^2 \cdot {}^n C_j. These are standard sums in the Binomial Theorem, and we often use specific identities to simplify their calculation. The key idea is to use the property jnCj=nn1Cj1j \cdot {}^n C_j = n \cdot {}^{n-1} C_{j-1} and j(j1)nCj=n(n1)n2Cj2j(j-1) \cdot {}^n C_j = n(n-1) \cdot {}^{n-2} C_{j-2} to transform the sums into simpler

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