Question

2+2+222222222222222​

Answer 1

Answer:

This might be easier for you. Take 2 common,. 2(1+11 +111+1111+11111+111111+1111111). Use this trick : Count how many digits are there in the last number, ...

Answer 2

Answer:

Consider the following composite series having n number of terms,

2+22+222+2222+…+(2222…n 2's)

=2⎛⎝1+11+111+1111+…+(1111…n 1's)⎞⎠

=29⎛⎝9+99+999+9999+⋯+(9999…n 9's)⎞⎠

=29((10−1)+(102−1)+(103−1)+(104−1)+⋯+(10n−1))

=29⎛⎝⎜⎜(10+102+103+…+10n)G.P. with n number of terms, first term 10 & common ratio 10−(1+1+1+1+…+1n 1's)⎞⎠⎟⎟

=29(10(10n−1)10−1−n)

=281(10(10n−1)−9n)

Hence, the generalized formula to get sum of n terms of this composite series

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Now, given series 2+22+222+2222+22222+222222+2222222 has seven terms hence substituting n=7 in the above generalized formula, the sum is given as

2+22+222+2222+22222+222222+2222222=281(10(107−1)−9⋅7)=2469134

Alternatively, one can fairly easily add all seven terms of given series as follows

2+22+222+2222+22222+222222+2222222––––––––––=2469134

NOTE: The most generalized formula to get sum of n terms of this composite series is given as

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Where, x is any digit i.e. x∈{0,1,2,3,4,5,6,7,8,9}