If (1+x)15=a0+a1x+...+a15x15, then ∑r=115r(arar-1) is
110
115
120
135
Explanation for the correct option:
Step 1: Expand (1+x)15
(1+x)15=150+151x+152x2+...+1515x15
Step 2: Comparing equation in step 1 with the given equation.
a0=150,a1=151,a2=152,.....,a15=1515
∴ar=15r
Step 3: Calculate rarar-1
rarar-1=r15r15r-1=r(15!)(15-r)!r!15!{15-(r-1)}!(r-1)!=r(15-r+1)r=15-r+1=16-r
Step 4: Calculate ∑r=115rarar-1
∑r=115rarar-1=∑r=115(16-r)=15+14+13+...+1=120
∴∑r=115rarar-1=120
Hence, option (C) is the correct option.
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