# If $$(10)^9$$ + $$2(11)^1(10)^8$$ + $$3(11)^2(10)^7$$ + …… + $$10(11)^9$$ = $$K(10)^9$$, then k is equal to

## Solution :

$$K(10)^9$$ = $$(10)^9$$ + $$2(11)^1(10)^8$$ + $$3(11)^2(10)^7$$ + …… + $$10(11)^9$$

K = 1 + 2$$({11\over 10})$$  + 3$$({11\over 10})^2$$ + ….. + 10$$({11\over 10})^9$$       ……(i)

$$({11\over 10})$$K = 1$$({11\over 10})$$ + 2$$({11\over 10})^2$$ + 3$$({11\over 10})^3$$ + ….. + 10$$({11\over 10})^{10}$$       …..(ii)

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

K$$(1 – {11\over 10})$$ = 1 + $$({11\over 10})$$ + $$({11\over 10})^2$$ + …. + $$({11\over 10})^9$$ – 10$$({11\over 10})^{10}$$

$$\implies$$ K$$({10 – 11\over 10})$$ = $$1[({11\over 10})^{10} – 1]\over ({11\over 10} – 1)$$ – 10$$({11\over 10})^{10}$$

$$\implies$$ – K = 10[10$$({11\over 10})^{10}$$ – 10 – 10$$({11\over 10})^{10}$$ ]

$$\implies$$ K = 100

### Similar Questions

Let $$a_n$$ be the nth term of an AP. If $$\sum_{r=1}^{100}$$ $$a_{2r}$$ = $$\alpha$$ and $$\sum_{r=1}^{100}$$ $$a_{2r-1}$$ = $$\beta$$, then the common difference of the AP is

The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ……. , is

If 100 times the 100th term of an AP with non-zero common difference equal to the 50 times its 50th term, then the 150th term of AP is

If x, y and z are in AP and $$tan^{-1}x$$, $$tan^{-1}y$$ and $$tan^{-1}z$$ are also in AP, then

Find the sum of n terms of the series 1.3.5 + 3.5.7 + 5.7.9 + ……