1
Question
The value of expression
3(1!)−4(2!)+5(3!)−6(4!)+.....−2008(2006!)+(2007!) is
The value of expression
3(1!)−4(2!)+5(3!)−6(4!)+.....−2008(2006!)+(2007!) is
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Solution
The correct option is C 1
T1=3(1!)=(2+1)(1!)=2!+1!T2=−(3+1)(2!)=−3!−2!T3=(4+1)(3!)=4!+3!T4=−(5+1)(4!)=−5!−4!...T2006=−(2007+1)(2006!)=−2007!−2006!
So the sum of all the terms will be ,
T1+T2+.......T2006=−2007!+1
So the value of expression =1
T1=3(1!)=(2+1)(1!)=2!+1!T2=−(3+1)(2!)=−3!−2!T3=(4+1)(3!)=4!+3!T4=−(5+1)(4!)=−5!−4!...T2006=−(2007+1)(2006!)=−2007!−2006!
So the sum of all the terms will be ,
T1+T2+.......T2006=−2007!+1
So the value of expression =1
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