Question

Find the sum of series: $1.2+{2.2}^2+{3.2}^3.....+{100.2}^{100}$

• A. ${199.2}^{100}+1$
• B. ${199.2}^{100}-2$
• C. ${199.2}^{100}+2$
• D. None of these

Answer 1

Answer: (D)

None of these

We know,

$(1+x{)}^{100}=1+x+x^2+x^3+...+x^{100}$

Differentiating w.r.t x, on both sides:

$100(1+x{)}^{99}=1+2x+3x^2+...+100x^{99}$

Multiplying with x on both sides

$100x(1+x{)}^{99}=1.x+2x^2+3x^3+...+100x^{100}$

Putting x=2,

$\left(100\right)\left(2\right){(1+x)}^{99}=1.2+2.2^2+3.2^3+...+100{\left(2\right)}^{100}$

$\left(200\right)3^{99}$ = The Given Series.