In these questions, the denominator contains the product of adjacent terms of an Arithmetic Progression. These can be solved by splitting each term and expression it as a difference of two terms.
Example 23
2×51+5×81+8×111+....+35×381=?
Solution
Note that 21−51=2×53 . Similarly 51−81=5×83 and so on.
We shall first express the terms with a numerator of 3.
2×51+5×81+8×111+....+35×381
=31×(2×53+5×83+8×113+......+35×383)
= 31×(21−51+51−81+81−111+......+351−381)
(All terms, other than the first and last terms will get cancelled.)
=31×(21−381)=31×3818=193
Answer: 193
5.5 Using sigma (Summation) and separating terms
The following formulae are commonly used to answer questions in CAT and other tests.
1) Sum of first n natural numbers = 2n(n+1)
2) Sum of squares of first n natural numbers = 6(n(n+1)(2n+1))
3) Sum of cubes of first n natural numbers = [2n(n+1)]2
4) Sum of first n even numbers = n(n+1)
5) Sum of first n odd numbers =n2
i=1∑nxi is a representation of “sum of all xi values, where i takes all integer values from 1 till n (both inclusive)”.
Therefore,i=1∑nxi=x1+x2+x3+.....+xn
Example 24
What is the sum of 1×5+2×6+3×7+.....+25×29 ?
Solution
Let x=1×5+2×6+3×7+.....+25×29
⇒ x=1×(1+4)+2×(2+4)+3×(3+4)+.....+25×(25+29)
⇒ x =n=1∑25n(n+4) = n=1∑25(n2+4n)
As this is just addition, we can separate the summation for n2 and 4n.
⇒ x =n=1∑25(n2) + n=1∑254n
A constant can be taken common outside the summation.
⇒ x =n=1∑25n2+(4×n=1∑25n)
We now apply the formula for sum of squares of n natural numbers and sum of n natural numbers.