Question

\[\text{If }{{a}_{k}}=\dfrac{1}{K\left( K+1 \right)}K=1,2,3,.....n,\text{ then }{{\left( \sum\limits_{k=1}^{n}{{{a}_{k}}} \right)}^{2}}=\]
(a) \[\dfrac{n}{n+1}\]
(b) \[\dfrac{{{n}^{2}}}{{{\left( n+1 \right)}^{2}}}\]
(c) \[\dfrac{{{n}^{4}}}{{{\left( n+1 \right)}^{4}}}\]
(d) \[\dfrac{{{n}^{6}}}{{{\left( n+1 \right)}^{6}}}\]

Answer 1

Hint: First try to simplify the term given. When we write it as a difference, we can use the concept of telescopic cancelation. For converting a product into a difference, we need to manipulate the numerator in terms of the denominator, then apply summation to the term. Then square this summation to get the required result.

Complete step-by-step answer:
Given, the term definition in the question is written as:
\[{{a}_{k}}=\dfrac{1}{K\left( K+1 \right)}\]
For this to convert into difference, we need the value of the numerator in terms of the expressions present in the denominator. The value of the numerator of the term is given by evaluation:
Numerator = 1
The algebraic expressions of the k which are in the denominator are given as:
K, K + 1
So, we can write the numerator in terms of these values as:
1 = (K + 1) – K
By this, we can say that the numerator will become:
Numerator = (K + 1) – K
By substituting this value into the term, we get it as:
\[{{a}_{k}}=\dfrac{\left( K+1 \right)-K}{K\left( K+1 \right)}\]
By general knowledge of the fractions, we know the relation as:
\[\dfrac{a-b}{c}=\dfrac{a}{c}-\dfrac{b}{c}\]
By substituting this fraction relation, we get the term as:
\[{{a}_{k}}=\dfrac{\left( K+1 \right)}{K\left( K+1 \right)}-\dfrac{K}{\left( K+1 \right)}\]
By canceling the common terms in both the fractions, we get it as,
\[{{a}_{k}}=\dfrac{1}{K}-\dfrac{1}{K+1}\]
By applying summation on both the sides of the evaluation, we get,
\[\sum\limits_{k-1}^{n}{{{a}_{k}}}=\sum\limits_{k=1}^{n}{\left[ \dfrac{1}{K}-\dfrac{1}{K+1} \right]}\]
By substituting the limits given, we get the summation as:
\[\sum\limits_{k-1}^{n}{{{a}_{k}}}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{n}-\dfrac{1}{n+1}\]
By canceling the common terms under the name telescopic cancellation, we get,
\[\sum\limits_{k-1}^{n}{{{a}_{k}}}=1-\dfrac{1}{n+1}\]
By simplifying the above equation, we get the equation as,
\[\sum\limits_{k-1}^{n}{{{a}_{k}}}=\dfrac{n+1-1}{n+1}=\dfrac{n}{n+1}\]
By squatting on both the sides, we get the result as
\[{{\left( \sum\limits_{k-1}^{n}{{{a}_{k}}} \right)}^{2}}={{\left( \dfrac{n}{n+1} \right)}^{2}}=\dfrac{{{n}^{2}}}{{{\left( n+1 \right)}^{2}}}\]
Hence, option (b) is the right answer.

Note: Be careful while using the concept of partial fraction, because the whole solution depends on the step of breaking. Generally students forget the minus sign in the fraction, this may lead to a great confusion, because you cannot cancel any terms in this position as there is a minus sign missing. Be careful while taking the least common multiple, as students confuse and write (n+1) + 1 instead of (n+1)-1 and this will change the whole answer.