Question

The sum of the infinite series `x + (1 + 2)/(2!) x^2 + (1 + 2 + 3)/(3!) x^3 +` .... equals

Options

• `1/2 x(x + 1)e^x`

• `1/2 x(x + 2)e^x`

• `1/2 x(2x + 1)e^x`

• `x(x + 2)e^x`

Answer 1

`1/2 x(x + 2)e^x`

Explanation:

Let `f(x) = 1/2 (x^2 + 2x) e^x` ⇒ `f(0)` = 0

`f"'"(x) = 1/2(2x + 2) e^x + 1/2 (x^2 + 2x) e^x = 1/2 (2x + 2) e^x + f(x)`

 ⇒ `f"'"(0)` = 1

`f"''"(x) = 1/2 . 2.e^x + 1/2 (2x + 2)e^x` + f"(x)

`f"''"(0)` = 1 + 1 + 1 = 1 + 2

By Taylor series.

`f(x) = f(0) + f"'"(0)`

`x + (f"''"(0))/(2!) x^2 + (f"''"(x))/(3!) x^3 + ... + (f"''"(x))/(n!) x^n + ...`

∴ `f(x) = 0 + 1.x + (1 + 2)/(2!) x^2 + (1 + 2 + 3)/(3!) x^2 + ...`

`1/2 x(x + 2)e^x = x + (1 + 2)/(2!) x^2 + (1 + 2 + 3)/(3!) x^3 + ...`