Find the sum of the series 5+55+555+... to n terms, let f(n) be Sn, then verify that S2=60 when n=2 using f(n), and lastly, establish that 81/x f(n) is always a multiple of 81

Do you know the equation for the nth term? You're right, these are pretty hard.

a = 5 tn=a(n-1)+5*10^(n-1) Does this seem right?

ahhhmmm, I've worked on it for so long, but there seems to be no progress

I dont know....

first term is 5*1 2nd= 5*12 3rd=5*123 4th=5*1234 Do you see the rule?

still I do not see how to change this into direct equation

yeah, maths HL in the IB diploma is unbelievably hard

what is HL and IB?

I am thinking of writing it up as a sum

\[5*\sum_{1}^{n} n10^{n-1}\]

Do you think it is right? :-)

I will post it again so someone can check it, ok?

HL is for higher level and IB is for International Bacchalerette

thanx

the answer is 5/81(10^(n+1)−9n−10)

check my topic if you need the calculation

thank you so much! but what do you mean by check my topics?

I reposted ur question

loved this problem btw! So thanks for posting

no, thank YOU for helping me with this! my maths teacher gaved these questions to me and told me that it took her a long time to do them too :-/

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