r=0∑10(10r1rr−1−1)11Cr+1=r=0∑10(10−10r1)11Cr+1=10r=0∑10Cr+1−10∑(11Cr+1(101)r+1)=10[11C1+11C2+…..+11C11]−10[11C1(101)1+11C2(101)2+…..+11C11(101)11]=10[211−1]−10[(1+101)11−1]=10(2)11−10−10101111+10=1010(20)11−1111∴α=20
JEE Main 2025 — Mathematics Algebra
If r=0∑10(10r10r+1−1)⋅11Cr+1=1010α11−1111, then α is equal to :
Held on 2 Apr 2025 · Verified 6 Jul 2026.
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