China
Find the sum of the sequence ^@5, 55, 555, 5555, ... ^@ to ^@ n ^@ terms.
Answer:
^@ \dfrac{ 5 } { 9 } \left[ \dfrac{ 10(10^n-1) }{9} -n \right] ^@
- The given sequence is not a ^@G.P.,^@ however, we can relate it to a ^@G.P.^@ by writing the given sequence differently.
- The sum of the given sequence up to ^@n^@ terms can be written as
^@\begin{align} S_n &= 5 + 55 + 555 + 5555 + \space .... \text{ to n terms}. \\ &= 5 [1 + 11 + 111 + 1111 + \space .... \text{ to n terms}] \\ &= \dfrac{ 5 } { 9 }[9 + 99 + 999 + 9999 + \space .... \text{ to n terms}] \\ &= \dfrac{ 5 } { 9 }[(10-1) + (10^2-1) + (10^3-1) + (10^4-1) + \space .... \text{ to n terms}] \\ &= \dfrac{ 5 } { 9 } \left[(10 + 10^2 + 10^3 + 10^4 + \space .... \text{ to n terms}) - (1+1+1+1 \space .... \text{ to n terms}) \right] \\ &= \dfrac{ 5 } { 9 } \left[ \dfrac{ 10(10^n-1) }{10-1} -n \right] \\ &= \dfrac{ 5 } { 9 } \left[ \dfrac{ 10(10^n-1) }{9} -n \right] \end{align}^@ - Hence, the sum of the sequence ^@5, 55, 555, 5555, ... ^@ to ^@ n ^@ terms is ^@ \dfrac{ 5 } { 9 } \left[ \dfrac{ 10(10^n-1) }{9} -n \right] ^@.
