Đáp án:
$\\$
`c,`
`A = 1 +5 + ... + 5^{49} + 5^{50}`
`-> 5A = 5 (1+5+...+5^{49} + 5^{50})`
`-> 5A= 5+5^2+...+5^{50} + 5^{51}`
`-> 5A-A = (5+5^2+...+5^{50} + 5^{51})- ( 1 +5 + ... + 5^{49} + 5^{50})`
`-> (5-1)A = 5^{51} - 1`
`-> 4A=5^{51}-1`
`->A=(5^{51}-1)/4`
Vậy `A=(5^{51}-1)/4`
$\\$
`d,`
`B = 1/2 + (1/2)^2 + ... + (1/2)^{98} + (1/2)^{99}`
`-> B = 1/2 + 1/2^2 + ... + 1/2^{98} + 1/2^{99}`
`-> 2B = 2 (1/2 + 1/2^2 + ... + 1/2^{98} +1/2^{99})`
`-> 2B = 1 + 1/2 + ... + 1/2^{97} + 1/2^{98}`
`-> 2B-B=(1 + 1/2 + ... + 1/2^{97} + 1/2^{98})-(1/2 + 1/2^2 + ... + 1/2^{98} + 1/2^{99})`
`-> (2-1)B = 1 - 1/2^{99}`
`-> B = 1 - 1/2^{99}`
Vậy `B = 1 - 1/2^{99}`
