Question

1/2+1/4+1/8+1/16+1/32+1/64+1/128=?

Answer 1

1/2;1/4;1/8;1/16;1/32;1/64;1/128 это геометрическое прогрессии
b1=1/2;q=1/2;|q|<1
S(7)=b1/(1-q)=1/2*1/1/2=1

Answer 2

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Answer 3

To find the sum of the given series, you can add up all the fractions:

1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128

To simplify the fractions and make the denominators the same, we notice that each fraction has a power of 2 in the denominator. We can rewrite all the fractions with a common denominator of 128 (2^7):

(64/128) + (32/128) + (16/128) + (8/128) + (4/128) + (2/128) + (1/128)

Now, we can add the fractions together:

(64 + 32 + 16 + 8 + 4 + 2 + 1) / 128

Combine the numerator:

127 / 128

So, the sum of the series 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 is 127/128.