Compute
\[\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \frac{1}{7 \times 10} + \dots + \frac{1}{97 \times 100}.\]
The given series is a geometric series with first term 1/4 and common ratio 3/4. The sum of a finite geometric series is given by the formula
S = a(1-r^n)/(1-r)
where a is the first term, r is the common ratio, and n is the number of terms. In this case, we have
S = 1/4(1-(3/4)^n)/(1-(3/4))
As n approaches infinity, the term (3/4)^n approaches 0, so the sum of the series approaches
S = 1/4/(1-(3/4)) = 1/3
Therefore, the sum of the given series is 1/3.
a=listfor(n, 1, 33, (1/(3*n - 2) * (1/(3*n+1)));print a; sum a
(1 / 4, 1 / 28, 1 / 70, 1 / 130, 1 / 208, 1 / 304, 1 / 418, 1 / 550, 1 / 700, 1 / 868, 1 / 1054, 1 / 1258, 1 / 1480, 1 / 1720, 1 / 1978, 1 / 2254, 1 / 2548, 1 / 2860, 1 / 3190, 1 / 3538, 1 / 3904, 1 / 4288, 1 / 4690, 1 / 5110, 1 / 5548, 1 / 6004, 1 / 6478, 1 / 6970, 1 / 7480, 1 / 8008, 1 / 8554, 1 / 9118, 1 / 9700) ==33/100
+397 The general term of the series can be taken to be 1/(n*(n + 3)) where n proceeds by 3 for each term, n = 1, 4, 7, 19, ... , 94, 97.
(It could also be taken to be 1/((3n - 2)(3n + 1)) where n proceeds in steps of 1 from 1 to 33, which was the choice of #2.)
Expanding 1/(n*(n + 3) ) into the partial fractions fractions 1/(3n) - 1/(3(n + 3)),
n = 1 : 1/(1*4) = 1/3 - 1/12,
n = 4 : 1/(4*7) = 1/12 - 1/21,
n = 7 : 1/(7*10) = 1/21 - 1/30,
n = 10: 1/(10*13) = 1/30 - 1/39,
.............
.............
n = 94 : 1/(94*97) = 1/282 - 1/291,
n = 97 : 1/(97*100) = 1/291 - 1/300.
Adding all of the equations,
1/(1*4) + 1/(4*7) +1/(7*10) + ... + 1/(97*100) = 1/3 - 1/300 = 100/300 - 1/300 = 99/300 = 33/100.
