Question

If (21)18 + 20·(21)17 + (20)2 · (21)16 + ……….. (20)18 = k (2119 – 2019) then k =

• A. $\frac{21}{20}$
• B. 1
• C. $\frac{20}{21}$
• D. 0

Answer 1

Answer: (B)

1

Answer 2

The correct answer is option (B): 1

$a=(21)^{18}.r=\frac{20}{21},n=19$

$S=(21)^{18}=\frac{(1-(\frac{20}{21})^{19})}{1-\frac{20}{21}}$

$\Rightarrow \frac{(21)^{19}}{(21)^{19}}(21^{19}-20^{19})$

$\Rightarrow(21^{19}-20^{19})=1$