Write an equation for 1 + 11 + 111 + 1111 ...?
f(n) = 1 + 11 + 111 + 1111 + ... where n is the number of 1's in the last term.
Write an equation for f(n) that doesn't use Epsilon or a series of terms. in your answer.
f(n) = 1 + 11 + 111 + 1111 + ... where n is the number of 1's in the last term.
Write an equation for f(n) that doesn't use Epsilon or a series of terms. in your answer.
Hemant
Favorite Answer
............ 1 + 11 + 111 + ... to (n) terms
= (1/9) [ 9 + 99 + 999 + ... to (n) terms ]
= (1/9) [ (10¹ - 1) + (10² - 1) + (10³ - 1) + ... + (10ⁿ - 1) ]
= (1/9) [ ( 10¹ + 10² + 10³ + ... + 10ⁿ ) - ( 1 + 1 + 1 + ... n times ) ]
= (1/9) { (10) [ ( 10ⁿ - 1 ) / (10-1) ] - n }
= (1/9) { (10/9) [ ( 10ⁿ - 1 ) ] - n } ........................ Ans.
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Anonymous
n=1
Robert Cotton
each term is
[(10^n)-1]/9
sum of n-1 terms plus nth term gives sum nth term so
s(n-1)+[(10^n)-1]/9=s(n)
likewise we know if we multiply sum n-1 times 10 and and n we get sum of the n terms so
10[s(n-1)]+n=s(n)
setting the two equal and solving for s(n-1)
10[s(n-1)]+n=s(n-1)+[(10^n)-1]/9
9[s(n-1)]={[(10^n)-1]/9}-n
9[s(n-1)]={[(10^n)-1]/9}-9n/9
9[s(n-1)]=[(10^n)-9n-1]/9
s(n-1)=[(10^n)-9n-1]/81
...
s(n)={10[(10^n)-9n-1]/81}+n
Anonymous
No, you do it! That ain't my ******* homework.