The value of (3⋅2⋅1P0−4⋅3⋅2P1+5⋅4⋅3P2−⋯upto101terms)+(2!−3!+4!−⋯upto101terms) is equal to
A
2!−102!
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B
2!−103!
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C
2!+103!
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D
3!+103!
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Solution
The correct option is C2!+103! Let S=(3⋅2⋅1P0−4⋅3⋅2P1+5⋅4⋅3P2−⋯upto101terms)+(2!−3!+4!−⋯upto101terms) ⇒S=(3⋅2⋅1P0−4⋅3⋅2P1+5⋅4⋅3P2−⋯+103⋅102⋅101P100)+(2!−3!+4!−⋯+102!)
Let S1=(3⋅2⋅1P0−4⋅3⋅2P1+5⋅4⋅3P2−⋯upto101terms) ⇒S1=3⋅2⋅1!1!−4⋅3⋅2!1!+5⋅4⋅3!1!−⋯+103⋅102⋅101!1! ⇒S1=3⋅2⋅1!−4⋅3⋅2!+5⋅4⋅3!−⋯+103⋅102⋅101! ⇒S1=3!−4!+5!−⋯+103!
Putting the value of S1 in S ∴S=(3!−4!+5!−⋯+103!)+(2!−3!+4!−5!+⋯+102!) ⇒S=2!+103!