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Find the sum (1xx2)/(3!)+(2xx2)^2/(4!)+(...

Find the sum `(1xx2)/(3!)+(2xx2)^2/(4!)+(3xx2)^3/(5!)+....+(20xx2)^30/(22!)`

Text Solution

Verified by Experts

The correct Answer is:
`1-(2^(21))/(22!)`
`T_(r)=(rxx2^(r ))/((r+2)!)=((r+2-2)cdot2^(r ))/((r+2)!)`
`=(2^(r ))/((r+1)!)-(2^(r+1))/((r+2)!)`
`therefore` Required sum=`sum_(r=1)^(20)[(2^(r ))/((r+1)!)-(2^(r+1))/((r+2)!)]`
`=2/(2!)-2^(21)/(22!)=1-2^(21)/(22!)`
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