
If ${t_n} = \dfrac{1}{4}(n + 2)(n + 3)$ for $n = 1,2,3.$ then $\dfrac{1}{{{t_1}}} + \dfrac{1}{{{t_2}}} + \dfrac{1}{{{t_3}}} + ....... + \dfrac{1}{{{t_{20}}}} = $?
Answer
554.7k+ views
Hint: First find the individual values of the first few terms and the last few terms, then start taking common values.
Initially obtain the values of few starting terms and few ending terms so that we will get some values and from further we will take common number so that we can form them into summable number i.e., into a more simplified and standard form
Complete step by step solution:
If ${t_n} = \dfrac{1}{4}(n + 2)(n + 3)$ for $n = 1,2,3.$
then $\dfrac{1}{{{t_1}}} + \dfrac{1}{{{t_2}}} + \dfrac{1}{{{t_3}}} + ....... + \dfrac{1}{{{t_{20}}}} = $
Now we have to separately find out the value of each single variable term and furthermore have to substitute the values into the given equation and start simplifying in more so that we can resolve the equation into as small as possible.
\[\
{t_1} = \dfrac{1}{4}(1 + 2)(1 + 3) \\
{t_1} = \dfrac{1}{4}(3)(4) \\
{t_2} = \dfrac{1}{4}(2 + 2)(2 + 3) \\
{t_2} = \dfrac{1}{4}(4)(5) \\
{t_3} = \dfrac{1}{4}(3 + 2)(3 + 3) \\
{t_3} = \dfrac{1}{4}(5)(6) \\
. \\
. \\
. \\
. \\
{t_{20}} = \dfrac{1}{4}(20 + 2)(20 + 3) \\
{t_{20}} = \dfrac{1}{4}(22)(23) \\
\]
Now, we have to solve the expression as told in the word-to-word explanation.
Further a take the common values aside and perform the calculation in the separate brackets so that the calculation gets simplified easily.
\[\
= \dfrac{1}{{{t_1}}} + \dfrac{1}{{{t_2}}} + \dfrac{1}{{{t_3}}} + ....... + \dfrac{1}{{{t_{20}}}} \\
= \dfrac{4}{{3 \times 4}} + \dfrac{4}{{4 \times 5}} + ............... + \dfrac{4}{{22 \times 23}} \\
= 4\left[ {\dfrac{1}{{3 \times 4}} + \dfrac{1}{{4 \times 5}} + ......... + \dfrac{1}{{22 \times 23}}} \right] \\
= 4\left[ {\dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{5} + ............. - \dfrac{1}{{23}}} \right] \\
= 4\left[ {\dfrac{1}{3} - \dfrac{1}{{23}}} \right] \\
= 4\left[ {\dfrac{{23 - 3}}{{69}}} \right] \\
= \dfrac{{80}}{{69}} \;
\]
So, the correct answer is “$\dfrac{{80}}{{69}}$”.
Note: When we come across problems like these, we have to first go with the brief reading of the problem and then the understanding of the question then thinking of the correct solution for the given question. The formulas play an important role in getting the equation solved. Without them solving the problem won’t be easy. In this problem we have to be careful while simplifying the equations and also in the fraction solving method make sure you write the correct coefficients so that you won’t be getting the solutions or the values.
Initially obtain the values of few starting terms and few ending terms so that we will get some values and from further we will take common number so that we can form them into summable number i.e., into a more simplified and standard form
Complete step by step solution:
If ${t_n} = \dfrac{1}{4}(n + 2)(n + 3)$ for $n = 1,2,3.$
then $\dfrac{1}{{{t_1}}} + \dfrac{1}{{{t_2}}} + \dfrac{1}{{{t_3}}} + ....... + \dfrac{1}{{{t_{20}}}} = $
Now we have to separately find out the value of each single variable term and furthermore have to substitute the values into the given equation and start simplifying in more so that we can resolve the equation into as small as possible.
\[\
{t_1} = \dfrac{1}{4}(1 + 2)(1 + 3) \\
{t_1} = \dfrac{1}{4}(3)(4) \\
{t_2} = \dfrac{1}{4}(2 + 2)(2 + 3) \\
{t_2} = \dfrac{1}{4}(4)(5) \\
{t_3} = \dfrac{1}{4}(3 + 2)(3 + 3) \\
{t_3} = \dfrac{1}{4}(5)(6) \\
. \\
. \\
. \\
. \\
{t_{20}} = \dfrac{1}{4}(20 + 2)(20 + 3) \\
{t_{20}} = \dfrac{1}{4}(22)(23) \\
\]
Now, we have to solve the expression as told in the word-to-word explanation.
Further a take the common values aside and perform the calculation in the separate brackets so that the calculation gets simplified easily.
\[\
= \dfrac{1}{{{t_1}}} + \dfrac{1}{{{t_2}}} + \dfrac{1}{{{t_3}}} + ....... + \dfrac{1}{{{t_{20}}}} \\
= \dfrac{4}{{3 \times 4}} + \dfrac{4}{{4 \times 5}} + ............... + \dfrac{4}{{22 \times 23}} \\
= 4\left[ {\dfrac{1}{{3 \times 4}} + \dfrac{1}{{4 \times 5}} + ......... + \dfrac{1}{{22 \times 23}}} \right] \\
= 4\left[ {\dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{5} + ............. - \dfrac{1}{{23}}} \right] \\
= 4\left[ {\dfrac{1}{3} - \dfrac{1}{{23}}} \right] \\
= 4\left[ {\dfrac{{23 - 3}}{{69}}} \right] \\
= \dfrac{{80}}{{69}} \;
\]
So, the correct answer is “$\dfrac{{80}}{{69}}$”.
Note: When we come across problems like these, we have to first go with the brief reading of the problem and then the understanding of the question then thinking of the correct solution for the given question. The formulas play an important role in getting the equation solved. Without them solving the problem won’t be easy. In this problem we have to be careful while simplifying the equations and also in the fraction solving method make sure you write the correct coefficients so that you won’t be getting the solutions or the values.
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