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Question: In the given series \(\dfrac{1}{2.5.8}+\dfrac{1}{5.8.11}+\dfrac{1}{8.11.14}+...+\text{up to}~n\text{...

In the given series 12.5.8+15.8.11+18.11.14+...+up to n terms.\dfrac{1}{2.5.8}+\dfrac{1}{5.8.11}+\dfrac{1}{8.11.14}+...+\text{up to}~n\text{ terms}.
Find the sum of n terms.

Explanation

Solution

To find the sum of n terms, first of all we will find the nth{{n}^{th}} term of the given equation. When we will get the nth{{n}^{th}} term of the given equation, we will use a summation sign in every term of the nth{{n}^{th}} term. Since, we used a summation sign that means we got the sum of the given equation up to n terms but if there is any possibility to simplify it, we will get the simplified sum of the given question up to n terms.

Complete step-by-step solution:
Since, the given question is:
⇒12.5.8+15.8.11+18.11.14+...+up to n terms.\Rightarrow \dfrac{1}{2.5.8}+\dfrac{1}{5.8.11}+\dfrac{1}{8.11.14}+...+\text{up to}~n\text{ terms}.
First of all we will get the nthterm{{n}^{th}}term of the given equation as:
⇒nthterm=1(3n−1)(3n+2)(3n+5)\Rightarrow {{n}^{th}}term=\dfrac{1}{\left( 3n-1 \right)\left( 3n+2 \right)\left( 3n+5 \right)}
Now, we will elaborate it by expanding the above equation. When we expand (3n−1)\left( 3n-1 \right) and (3n+5)\left( 3n+5 \right) so that we will get 66 as a result. So, there we will divide by 33 in the above equation as:
⇒nthterm=1(3n+2)16(1(3n−1)−1(3n+5))\Rightarrow {{n}^{th}}term=\dfrac{1}{\left( 3n+2 \right)}\dfrac{1}{6}\left( \dfrac{1}{\left( 3n-1 \right)}-\dfrac{1}{\left( 3n+5 \right)} \right)
After that we will open the bracket and will multiply (3n−1)\left( 3n-1 \right)to the terms that are inside the bracket as:
⇒nthterm=16(1(3n−1)(3n+2)−1(3n+2)(3n+5))\Rightarrow {{n}^{th}}term=\dfrac{1}{6}\left( \dfrac{1}{\left( 3n-1 \right)\left( 3n+2 \right)}-\dfrac{1}{\left( 3n+2 \right)\left( 3n+5 \right)} \right)
Now, again we will separate the in bracketed terms that are in multiplication form as:
⇒nthterm=13[(1(3n−1)−1(3n+2))−(1(3n+2)−1(3n+5))]\Rightarrow {{n}^{th}}term=\dfrac{1}{3}\left[ \left( \dfrac{1}{\left( 3n-1 \right)}-\dfrac{1}{\left( 3n+2 \right)} \right)-\left( \dfrac{1}{\left( 3n+2 \right)}-\dfrac{1}{\left( 3n+5 \right)} \right) \right]
Since, we will get 33when we will use subtraction in both the bracketed terms. So we will divide them by 33 as:
⇒nthterm=16[13(1(3n−1)−1(3n+2))−13(1(3n+2)−1(3n+5))]\Rightarrow {{n}^{th}}term=\dfrac{1}{6}\left[ \dfrac{1}{3}\left( \dfrac{1}{\left( 3n-1 \right)}-\dfrac{1}{\left( 3n+2 \right)} \right)-\dfrac{1}{3}\left( \dfrac{1}{\left( 3n+2 \right)}-\dfrac{1}{\left( 3n+5 \right)} \right) \right]
Now, we will take out 13\dfrac{1}{3} as a common factor and will multiply it by 16\dfrac{1}{6} as:
⇒nthterm=16×13[(1(3n−1)−1(3n+2))−(1(3n+2)−1(3n+5))]\Rightarrow {{n}^{th}}term=\dfrac{1}{6}\times \dfrac{1}{3}\left[ \left( \dfrac{1}{\left( 3n-1 \right)}-\dfrac{1}{\left( 3n+2 \right)} \right)-\left( \dfrac{1}{\left( 3n+2 \right)}-\dfrac{1}{\left( 3n+5 \right)} \right) \right]
⇒nthterm=118[(1(3n−1)−1(3n+2))−(1(3n+2)−1(3n+5))]\Rightarrow {{n}^{th}}term=\dfrac{1}{18}\left[ \left( \dfrac{1}{\left( 3n-1 \right)}-\dfrac{1}{\left( 3n+2 \right)} \right)-\left( \dfrac{1}{\left( 3n+2 \right)}-\dfrac{1}{\left( 3n+5 \right)} \right) \right]
Here, we will solve the bracketed terms by opening the bracket as:
⇒nthterm=118[1(3n−1)−1(3n+2)−1(3n+2)+1(3n+5)]\Rightarrow {{n}^{th}}term=\dfrac{1}{18}\left[ \dfrac{1}{\left( 3n-1 \right)}-\dfrac{1}{\left( 3n+2 \right)}-\dfrac{1}{\left( 3n+2 \right)}+\dfrac{1}{\left( 3n+5 \right)} \right]
Now, the above equation will be as:
⇒nthterm=118[1(3n−1)−2(3n+2)+1(3n+5)]\Rightarrow {{n}^{th}}term=\dfrac{1}{18}\left[ \dfrac{1}{\left( 3n-1 \right)}-\dfrac{2}{\left( 3n+2 \right)}+\dfrac{1}{\left( 3n+5 \right)} \right]
Now, here we will try to find out the value of ever term as:
For first term: n=1n=1
⇒1stterm=118[1(3×1−1)−2(3×1+2)+1(3×1+5)]\Rightarrow {{1}^{st}}term=\dfrac{1}{18}\left[ \dfrac{1}{\left( 3\times 1-1 \right)}-\dfrac{2}{\left( 3\times 1+2 \right)}+\dfrac{1}{\left( 3\times 1+5 \right)} \right]
⇒1stterm=118[1(3−1)−2(3+2)+1(3+5)]\Rightarrow {{1}^{st}}term=\dfrac{1}{18}\left[ \dfrac{1}{\left( 3-1 \right)}-\dfrac{2}{\left( 3+2 \right)}+\dfrac{1}{\left( 3+5 \right)} \right]
⇒1stterm=118[12−25+18]\Rightarrow {{1}^{st}}term=\dfrac{1}{18}\left[ \dfrac{1}{2}-\dfrac{2}{5}+\dfrac{1}{8} \right]
For second term: n=2n=2
⇒2ndterm=118[1(3×2−1)−2(3×2+2)+1(3×2+5)]\Rightarrow {{2}^{nd}}term=\dfrac{1}{18}\left[ \dfrac{1}{\left( 3\times 2-1 \right)}-\dfrac{2}{\left( 3\times 2+2 \right)}+\dfrac{1}{\left( 3\times 2+5 \right)} \right]
After solving above equation we will get:
⇒2ndterm=118[15−28+111]\Rightarrow {{2}^{nd}}term=\dfrac{1}{18}\left[ \dfrac{1}{5}-\dfrac{2}{8}+\dfrac{1}{11} \right]
For third term: n=3n=3
⇒3rdterm=118[1(3×3−1)−2(3×3+2)+1(3×3+5)]\Rightarrow {{3}^{rd}}term=\dfrac{1}{18}\left[ \dfrac{1}{\left( 3\times 3-1 \right)}-\dfrac{2}{\left( 3\times 3+2 \right)}+\dfrac{1}{\left( 3\times 3+5 \right)} \right]
We will have the value of third term after solving the above equation as:
⇒3rdterm=118[18−211+114]\Rightarrow {{3}^{rd}}term=\dfrac{1}{18}\left[ \dfrac{1}{8}-\dfrac{2}{11}+\dfrac{1}{14} \right]
… … … … …
For (n−1)thterm{{\left( n-1 \right)}^{th}}term : n=n−1n=n-1
⇒(n−1)thterm=118[13(n−1)−1−23(n−1)+2+13(n−1)+5]\Rightarrow {{\left( n-1 \right)}^{th}}term=\dfrac{1}{18}\left[ \dfrac{1}{3\left( n-1 \right)-1}-\dfrac{2}{3\left( n-1 \right)+2}+\dfrac{1}{3\left( n-1 \right)+5} \right]
Now, the (n−1)thterm{{\left( n-1 \right)}^{th}}term will be as:
⇒(n−1)thterm=118[13n−4−23n−1+13n+2]\Rightarrow {{\left( n-1 \right)}^{th}}term=\dfrac{1}{18}\left[ \dfrac{1}{3n-4}-\dfrac{2}{3n-1}+\dfrac{1}{3n+2} \right]
This process will go up to nthterm{{n}^{th}}term.
Since, we got all the n termsn\text{ }terms of the given question. Now, we will apply summation sign to get the sum of the given question as:
⇒Sn=118[∑1n1(3n−1)−∑1n2(3n+2)+∑1n1(3n+5)]\Rightarrow {{S}_{n}}=\dfrac{1}{18}\left[ \sum\limits_{1}^{n}{\dfrac{1}{\left( 3n-1 \right)}}-\sum\limits_{1}^{n}{\dfrac{2}{\left( 3n+2 \right)}}+\sum\limits_{1}^{n}{\dfrac{1}{\left( 3n+5 \right)}} \right]
Now, we will write it as:
⇒Sn=118[12−25+18]+118[15−28+111]+118[18−211+114]+... +118[13n−4−23n−1+13n+2]+118[1(3n−1)−2(3n+2)+1(3n+5)] \begin{aligned} & \Rightarrow {{S}_{n}}=\dfrac{1}{18}\left[ \dfrac{1}{2}-\dfrac{2}{5}+\dfrac{1}{8} \right]+\dfrac{1}{18}\left[ \dfrac{1}{5}-\dfrac{2}{8}+\dfrac{1}{11} \right]+\dfrac{1}{18}\left[ \dfrac{1}{8}-\dfrac{2}{11}+\dfrac{1}{14} \right]+... \\\ & +\dfrac{1}{18}\left[ \dfrac{1}{3n-4}-\dfrac{2}{3n-1}+\dfrac{1}{3n+2} \right]+\dfrac{1}{18}\left[ \dfrac{1}{\left( 3n-1 \right)}-\dfrac{2}{\left( 3n+2 \right)}+\dfrac{1}{\left( 3n+5 \right)} \right] \\\ \end{aligned}
⇒Sn=118[12−25+18+15−28+111+18−211+114+...+13n−4−23n−1+13n+2+1(3n−1)−2(3n+2)+1(3n+5)]\Rightarrow {{S}_{n}}=\dfrac{1}{18}\left[ \dfrac{1}{2}-\dfrac{2}{5}+\dfrac{1}{8}+\dfrac{1}{5}-\dfrac{2}{8}+\dfrac{1}{11}+\dfrac{1}{8}-\dfrac{2}{11}+\dfrac{1}{14}+...+\dfrac{1}{3n-4}-\dfrac{2}{3n-1}+\dfrac{1}{3n+2}+\dfrac{1}{\left( 3n-1 \right)}-\dfrac{2}{\left( 3n+2 \right)}+\dfrac{1}{\left( 3n+5 \right)} \right]
As we can see that some terms will be canceled. So, we will get the above equation as:
⇒Sn=118[12−15−1(3n+2)+1(3n+5)]\Rightarrow {{S}_{n}}=\dfrac{1}{18}\left[ \dfrac{1}{2}-\dfrac{1}{5}-\dfrac{1}{\left( 3n+2 \right)}+\dfrac{1}{\left( 3n+5 \right)} \right]
Now, we will solve the above equation as:
⇒Sn=118[5−210−((3n+5)−(3n+2)(3n+2)(3n+5))]\Rightarrow {{S}_{n}}=\dfrac{1}{18}\left[ \dfrac{5-2}{10}-\left( \dfrac{\left( 3n+5 \right)-\left( 3n+2 \right)}{\left( 3n+2 \right)\left( 3n+5 \right)} \right) \right]
⇒Sn=118[310−3(3n+2)(3n+5)]\Rightarrow {{S}_{n}}=\dfrac{1}{18}\left[ \dfrac{3}{10}-\dfrac{3}{\left( 3n+2 \right)\left( 3n+5 \right)} \right]
Here, we will take 33 as common:
⇒Sn=318[110−1(3n+2)(3n+5)]\Rightarrow {{S}_{n}}=\dfrac{3}{18}\left[ \dfrac{1}{10}-\dfrac{1}{\left( 3n+2 \right)\left( 3n+5 \right)} \right]
Since, we got a sum of the n terms of the given question but we can simplify it further. So, we will use subtraction to solve it as:
⇒Sn=16[(3n+2)(3n+5)−1010(3n+2)(3n+5)]\Rightarrow {{S}_{n}}=\dfrac{1}{6}\left[ \dfrac{\left( 3n+2 \right)\left( 3n+5 \right)-10}{10\left( 3n+2 \right)\left( 3n+5 \right)} \right]
Now, we will expand the above equation by opening bracket as:
⇒Sn=16[9n2+6n+15n+10−1010(9n2+6n+15n+10)]\Rightarrow {{S}_{n}}=\dfrac{1}{6}\left[ \dfrac{9{{n}^{2}}+6n+15n+10-10}{10\left( 9{{n}^{2}}+6n+15n+10 \right)} \right]
Now, we will do the necessary calculation as:
⇒Sn=16[9n2+21n(90n2+210n+100)]\Rightarrow {{S}_{n}}=\dfrac{1}{6}\left[ \dfrac{9{{n}^{2}}+21n}{\left( 90{{n}^{2}}+210n+100 \right)} \right]
⇒Sn=16[3(3n2+7n)(90n2+210n+100)]\Rightarrow {{S}_{n}}=\dfrac{1}{6}\left[ \dfrac{3\left( 3{{n}^{2}}+7n \right)}{\left( 90{{n}^{2}}+210n+100 \right)} \right]
⇒Sn=36[n(3n+7)(90n2+210n+100)]\Rightarrow {{S}_{n}}=\dfrac{3}{6}\left[ \dfrac{n\left( 3n+7 \right)}{\left( 90{{n}^{2}}+210n+100 \right)} \right]
⇒Sn=12[n(3n+7)(90n2+210n+100)]\Rightarrow {{S}_{n}}=\dfrac{1}{2}\left[ \dfrac{n\left( 3n+7 \right)}{\left( 90{{n}^{2}}+210n+100 \right)} \right]
Hence, this is a simplified solution of the sum of n terms of the given equation.

Note: Here, we will use alternative method to check the sum of the given question as:
Since, we will have to find the sum of the given question for infinite terms that is:
⇒12.5.8+15.8.11+18.11.14+...\Rightarrow \dfrac{1}{2.5.8}+\dfrac{1}{5.8.11}+\dfrac{1}{8.11.14}+...
Here, we will consider the sum of some terms as:
⇒12.5+15.8+18.11+... = S∞\Rightarrow \dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...\text{ = }{{S}_{\infty }} … (i)\left( i \right)
Now, we can change the place of first term and can place them other side of equal sign as:
⇒15.8+18.11+... = S∞−12.5\Rightarrow \dfrac{1}{5.8}+\dfrac{1}{8.11}+...\text{ = }{{S}_{\infty }}-\dfrac{1}{2.5} … (ii)\left( ii \right)
Since, we have two equations; we need to subtract equation (i)\left( i \right) from equation (ii)\left( ii \right) so that we can solve the question easily as:

& \underline{\begin{aligned} & \Rightarrow \dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...\text{ = }{{S}_{\infty }} \\\ & \Rightarrow \dfrac{1}{5.8}+\dfrac{1}{8.11}+...\text{ = }{{S}_{\infty }}-\dfrac{1}{2.5} \\\ \end{aligned}} \\\ & \Rightarrow \dfrac{6}{2.5.8}+\dfrac{6}{5.8.11}+\dfrac{6}{8.11.14}...\text{ =}\dfrac{1}{2.5} \\\ \end{aligned}$$ Since, $6$ is a common factor in all the terms of the above equation. So, we can take $6$common as: $$\Rightarrow 6\left( \dfrac{1}{2.5.8}+\dfrac{1}{5.8.11}+\dfrac{1}{8.11.14}+... \right)\text{ =}\dfrac{1}{2.5}$$ As we have $6$ , a common factor in the above equation. So we can divide by $6$ in the above equation as: $$\Rightarrow \dfrac{6}{6}\left( \dfrac{1}{2.5.8}+\dfrac{1}{5.8.11}+\dfrac{1}{8.11.14}+... \right)\text{ =}\dfrac{1}{2.5}\times \dfrac{1}{6}$$ Here, $6$ will cancel out $6$ and after that we will have the multiplication of $2$ , $5$ and $6$as $$\Rightarrow \left( \dfrac{1}{2.5.8}+\dfrac{1}{5.8.11}+\dfrac{1}{8.11.14}+... \right)\text{ =}\dfrac{1}{60}$$ Hence, we have $\dfrac{1}{60}$ as the sum of infinite terms of a given question. Since, from the solution we have ${{n}^{th}}term$ of the given question as: $\Rightarrow {{n}^{th}}term=\dfrac{1}{18}\left[ \dfrac{1}{\left( 3n-1 \right)}-\dfrac{2}{\left( 3n+2 \right)}+\dfrac{1}{\left( 3n+5 \right)} \right]$ And the sum of the solution is: $\Rightarrow {{S}_{\infty }}=\dfrac{1}{18}\left[ \dfrac{3}{10}-\dfrac{3}{\left( 3n+2 \right)\left( 3n+5 \right)} \right]$ Since, we will replace n by infinite. So, the second term of the bracket approximately will be equal to zero. Hence we will have: $\Rightarrow {{S}_{\infty }}=\dfrac{1}{18}\left[ \dfrac{3}{10}-0 \right]$ Now, we will solve the above equation as: $\Rightarrow {{S}_{\infty }}=\dfrac{1}{18}\times \dfrac{3}{10}$ $\Rightarrow {{S}_{\infty }}=\dfrac{1}{6}\times \dfrac{1}{10}$ $\Rightarrow {{S}_{\infty }}=\dfrac{1}{60}$ Here, we also get the value of the sum of infinite terms of a given question as $\dfrac{1}{60}$ . Hence, the solution is correct.