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Find the sum of the series 4+42+43+....4100+4.4101
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Q
2
sin−1x+sin−1y+sin−1z=3π2, then the value of x100+y100+z100−9x101+y101+z101=
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Coefficient of x100in1+(1+x)+(1+x)2+(1+x)3…(1+x)n is 201C101,(n>100) then n =
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Q
4
(1×103)+(2×102)+(3×101)+(4×100)+(2×10−1)+(3×10−2) =
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Let Sn=1+q+q2+⋯+qn and Tn=1+(q+12)+(q+12)2+⋯+(q+12)n where q is a real number and q≠1. If 101C1+101C2⋅S1+⋯+ 101C101⋅S100=α T100, then α is equal vĩ đại :
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