The value of \(\frac{1}{3\times7}+\frac{1}{7\times11}+\frac{1}{11\times15}+\ldots+\frac{1}{27\times31}\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\frac{7}{93}\)
In every term the two factors of the denominator differ by 4, so each term splits into a difference of two fractions, and the middle fractions cancel in pairs (a telescoping sum).
\(\displaystyle \frac{1}{a(a+4)}=\frac{1}{4}\left(\frac{1}{a}-\frac{1}{a+4}\right)\)
Sum \(\displaystyle =\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\ldots+\frac{1}{27}-\frac{1}{31}\right)\)
Only the first and last fractions are left: \(\displaystyle \frac{1}{4}\left(\frac{1}{3}-\frac{1}{31}\right)=\frac{1}{4}\times\frac{28}{93}\)
\(\displaystyle =\frac{7}{93}\)
Tip: Leaving out the factor \(\frac{1}{4}\) gives \(\frac{28}{93}\), which is Option (1). The factor is 1 divided by the gap between the two factors (here 4).