Find the sum of the first 16 terms.

Given series:

1/(4×10) + 1/(10×16) + 1/(16×22) + …

General term:

Tn = 1/[(6n − 2)(6n + 4)], since n = 1 gives 1/(4·10), n = 2 gives 1/(10·16), etc.

Use partial fractions to decompose Tn:

• 1/[(6n − 2)(6n + 4)] = (1/6)·(1/(6n − 2) − 1/(6n + 4)).

Therefore the sum of the first 16 terms is

S16 = (1/6)·[ (1/4 − 1/10) + (1/10 − 1/16) + (1/16 − 1/22) + … + (1/94 − 1/100) ].

This is a telescoping sum: most intermediate terms cancel, leaving 1/4 − 1/100.

So S16 = (1/6)·(1/4 − 1/100) = (1/6)·(25/100 − 1/100) = (1/6)·(24/100) = 24/600 = 1/25.

Final answer: 1/25.