Sum of the series nC1+2⋅5nC2+3⋅52nC3+⋯ upto n terms is
A
n⋅6n−1
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B
6n−1
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C
6n
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D
n⋅6n
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Solution
The correct option is An⋅6n−1 Let S=nC1+2⋅5nC2+3⋅52nC3+⋯ upto n terms =n∑r=15r−1⋅r⋅nCr =n∑r=15r−1⋅n⋅n−1Cr−1 =nn∑r=1n−1Cr−1⋅5r−1 =n⋅(1+5)n−1(∵n∑s=0nCs⋅ar=(1+a)n) =n⋅6n−1