The sum 1+2·3+3·3^2+……..+10·3^9 is equal to

Correct answer: (B) (19·3^10+1)/4. Let S=1·3^0+2·3^1+3·3^2+…..+10.3^9 ....(i) 3S=1.3^1+2.3^2……………….+9×3^9+10×3^10 .....(ii) Now subtracting equation (ii) from (i) we get, -2 S=1·3^0+3^1+3^2…..3^9-10.3^10 S=5×3^10-(3^10-1)/4 S=(20.3^10-3^10+1)/4=(19.3^10+1)/4

Q38 / 90 · JEE Main 2022 (25 Jun Shift 2) · Mathematics

The sum $1 + 2 \cdot 3 + 3 \cdot 3^{2} + \dots \dots . . + 10 \cdot 3^{9}$ is equal to

A
$\frac{2 \cdot 3^{12} + 10}{4}$
Correct answer
B
$\frac{19 \cdot 3^{10} + 1}{4}$
C
$5 \cdot 3^{10} - 2$
D
$\frac{9 \cdot 3^{10} + 1}{2}$

Solution

Let $S = 1 \cdot 3^{0} + 2 \cdot 3^{1} + 3 \cdot 3^{2} + \dots . . + 10 . 3^{9}$ ....(i)

$3 S = 1 . 3^{1} + 2 . 3^{2} \dots \dots \dots \dots \dots \dots . + 9 \times 3^{9} + 10 \times 3^{10}$ .....(ii)

Now subtracting equation (ii) from (i) we get,

$- 2 S = (1 \cdot 3^{0} + 3^{1} + 3^{2} \dots . . 3^{9}) - 10 . 3^{10}$

$S = 5 \times 3^{10} - (\frac{3^{10} - 1}{4})$

$S = \frac{20 . 3^{10} - 3^{10} + 1}{4} = \frac{19 . 3^{10} + 1}{4}$

Hence, the correct answer is option: 'B'.

The sum 1+2·3+3·32+……..+10·39 is equal to

Solution

Relevant Questions

The sum $1 + 2 \cdot 3 + 3 \cdot 3^{2} + \dots \dots . . + 10 \cdot 3^{9}$ is equal to