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The sum of the series (4)/(1!)+(11)/(2!)...

The sum of the series `(4)/(1!)+(11)/(2!)+(22)/(3!)+(37)/(4!)+(56)/(5!)`+..is

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Let us first ` n^(th)` term `(t_(n)) ` or the series 4, 11, 22, 37 , ….
`S_(n) = 4 + 11 + 22 + 37 + 56 +… + t_(n) `
` S_(n) = 4 + 11+ 22 + 37 + …+ t_(n-1) + t_(n) `
Subtract ` 0 = 4 + 7 + 11 + 15 + 19 + … (t_(n) - t_(n-1) ) - t_(n)`
or ` t_(n) = 4 + 7 + 11 + 15 + 19 + ... "upto" n^(th)` term
= ` 4 + (7 + 11 + 15 + 19 + ...` upto (n-1) term)
` = 4 + (n-1)/(2) [ 14 + (n-2) 4]`
` = 4 + (n-1) [ 2n + 3] `
` 2n^(2) + n+1`
Hence ` n^(th)` term of given series is
` T_(n) = (2n^(2) + n+ 1)/(n!)`
` = (2n)/((n-1)!) + (1)/((n-1)!) + (1)/(n!)`
`= (2)/((n-2)!) + (3)/((n-1)!)+(1)/(n!)`
` therefore S_(n) = s sum_(n=2)^(oo) [(1)/((n-2)!)] + 3 sum_(n=1)^(oo) (1)/((n-1)!) + sum_(n=1)^(oo) (1)/(n!)`
` = 2e + 3e + (e - 1)`
` 6 e - 1`
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