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Algebra Level 3

4 + 44 + 444 + ⋯ + 44444 … 4 ⏟ n 4’s 4 + 44 + 444 + \cdots + \underbrace{44444\ldots4}_{n\text{ 4's}}

Find the closed form of the summation above.

4 81 [ 10 ( 1 0 n + 1 ) − 9 n ] \dfrac4{81}[10(10^n+1)-9n] 4 81 [ 10 ( 1 0 n + 1 ) + 9 n ] \dfrac4{81}[10(10^n+1)+9n] 4 81 [ 10 ( 1 0 n − 1 ) − 9 n ] \dfrac4{81}[10(10^n-1)-9n] 4 81 [ 10 ( 1 0 n − 1 ) + 9 n ] \dfrac4{81}[10(10^n-1)+9n]

Sabhrant Sachan by Sabhrant Sachan , 21, India

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2 solutions

Sabhrant Sachan
May 1, 2016

Relevant wiki: Arithmetic and Geometric Progressions Problem Solving

Let , S n = 4 + 44 + 444 + 4444 + . . . . (n terms) ⟹ S n = 4 ( 1 + 11 + 111 + 1111 + . . . . n terms ) ⟹ S n = 4 9 ( 9 + 99 + 999 + 9999 + . . . . n terms ) ⟹ S n = 4 9 [ ( 10 − 1 ) + ( 100 − 1 ) + ( 1000 − 1 ) + ( 10000 − 1 ) + . . . . n terms ] ⟹ S n = 4 9 [ ( 10 + 1 0 2 + 1 0 3 + ⋯ n terms ) − ( 1 + 1 + 1 + ⋯ n terms ) ] S n = 4 9 ( 10 ( 1 0 n − 1 ) 10 − 1 − n ) S n = 4 81 [ 10 ( 1 0 n − 1 ) − 9 n ] \text {Let , } S_n=4+44+444+4444+....\text {(n terms)} \\ \implies S_n=4(1+11+111+1111+....\text {n terms}) \\ \implies S_n=\dfrac49(9+99+999+9999+....\text {n terms}) \\ \implies S_n=\dfrac49[(10-1)+(100-1)+(1000-1)+(10000-1)+....\text {n terms}] \\ \implies S_n=\dfrac49[(10+10^2+10^3+\cdots\text{n terms})-(1+1+1+\cdots\text{n terms})] \\ S_n=\dfrac49(\dfrac{10(10^n-1)}{10-1}-n)\\ S_n=\dfrac{4}{81}[10(10^n-1)-9n]

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Sayandeep Ghosh
May 4, 2016

I have approached like this solution only....

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