
Sum of the series \[4+6+9+13+18+..........\]up to \['n'\] terms be \[\dfrac{n}{k}\left( {{n}^{2}}+3n+m \right)\] then find the value of \[m-k\]
Answer
465.6k+ views
Hint: We solve this problem first by finding the \[{{n}^{th}}\] term. Then we use the standard formula of sum of n terms using the \[{{n}^{th}}\] term that is
\[{{S}_{n}}=\sum{{{T}_{n}}}\]
We use some standard results of sum of terms that is
\[\sum{n}=\dfrac{n\left( n+1 \right)}{2}\]
\[\sum{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]
\[\sum{1}=n\]
By using the above results we find the sum of terms to get the required value.
Complete step by step answer:
We are given that the series as \[4+6+9+13+18+..........\]up to \['n'\] terms
Let us assume that the sum of the given series as
\[\Rightarrow {{S}_{n}}=4+6+9+13+.............+{{T}_{n-1}}+{{T}_{n}}...........equation(i)\]
Now let us add ‘0’ on both sides so that the result will not change that is
\[\Rightarrow {{S}_{n}}=0+4+6+9+13+............+{{T}_{n-1}}+{{T}_{n}}...........equation(ii)\]
By subtracting the equation (ii) from equation (i) we get
\[\begin{align}
& \Rightarrow 0=4+\left( 2+3+4+...........\left( n-1 \right)terms \right)-{{T}_{n}} \\
& \Rightarrow {{T}_{n}}-4=2+3+4+...........\left( n-1 \right)terms \\
\end{align}\]
Here we can see that the RHS is in A.P
We know that the sum of \['n'\] terms in an A.P having \['a'\] as first term and \['d'\] as common difference is given as
\[\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\]
By using this formula to above equation we get
\[\begin{align}
& \Rightarrow {{T}_{n}}-4=\dfrac{n-1}{2}\left( 2\times 2+\left( n-1-1 \right)\times 1 \right) \\
& \Rightarrow {{T}_{n}}-4=\dfrac{\left( n-1 \right)\left( n+2 \right)}{2} \\
\end{align}\]
Now, by multiplying the terms on RHS we get
\[\begin{align}
& \Rightarrow {{T}_{n}}=\dfrac{1}{2}\left( {{n}^{2}}+n-2 \right)+4 \\
& \Rightarrow {{T}_{n}}=\dfrac{{{n}^{2}}}{2}+\dfrac{n}{2}+3 \\
\end{align}\]
We know that the standard formula of sum of n terms using the \[{{n}^{th}}\] term that is
\[{{S}_{n}}=\sum{{{T}_{n}}}\]
By using the above formula we get the sum of given series as
\[\Rightarrow {{S}_{n}}=\dfrac{1}{2}\left( \sum{{{n}^{2}}}+\sum{n} \right)+3\sum{1}.......equation(iii)\]
We know that the some of the standard results of sum of terms that is
\[\sum{n}=\dfrac{n\left( n+1 \right)}{2}\]
\[\sum{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]
\[\sum{1}=n\]
By using these formulas in equation (iii) we get
\[\Rightarrow {{S}_{n}}=\dfrac{1}{2}\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{n\left( n+1 \right)}{2} \right)+3n\]
Now by taking the common terms out we get
\[\begin{align}
& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( \dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{\left( n+1 \right)}{2} \right)+3n \\
& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( \dfrac{2{{n}^{2}}+3n+1}{6}+\dfrac{\left( n+1 \right)}{2}+6 \right) \\
\end{align}\]
Now, by using the LCM method and adding the terms we get
\[\begin{align}
& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( \dfrac{2{{n}^{2}}+3n+1+3n+3+36}{6} \right) \\
& \Rightarrow {{S}_{n}}=\dfrac{n}{2\times 6}\left( 2{{n}^{2}}+6n+40 \right) \\
& \Rightarrow {{S}_{n}}=\dfrac{n}{6}\left( {{n}^{2}}+3n+20 \right) \\
\end{align}\]
We are given that the sum of series as
\[\Rightarrow {{S}_{n}}=\dfrac{n}{k}\left( {{n}^{2}}+3n+m \right)\]
By comparing this given formula with the result we get
\[\begin{align}
& \Rightarrow m=20 \\
& \Rightarrow k=6 \\
\end{align}\]
Now by finding the value of \[m-k\] we get
\[\Rightarrow m-k=20-6=14\]
Therefore, the value of \[m-k\] is 14.
Note: We can solve this problem in other methods also.
We have the series as
\[4+6+9+13+18+..........\]
Here, we can see that the differences of consecutive terms are in A.P then the \[{{n}^{th}}\] term is given as
\[\Rightarrow {{T}_{n}}=a{{n}^{2}}+bn+c\]
By taking \[n=1\] we get
\[\Rightarrow a+b+c=4......equation(i)\]
By taking \[n=2\] we get
\[\Rightarrow 4a+2b+c=6......equation(ii)\]
By taking \[n=3\] we get
\[\Rightarrow 9a+3b+c=9......equation(iii)\]
Now by subtracting equation (i) from equation (ii) we get
\[\begin{align}
& \Rightarrow 3a+b=2 \\
& \Rightarrow b=2-3a \\
\end{align}\]
Now, by subtracting equation (ii) from equation (iii) we get
\[\begin{align}
& \Rightarrow 5a+b=3 \\
& \Rightarrow 5a+2-3a=3 \\
& \Rightarrow a=\dfrac{1}{2} \\
\end{align}\]
By substituting the value of \['a'\] in \['b'\] we get
\[\begin{align}
& \Rightarrow b=2-\dfrac{3}{2} \\
& \Rightarrow b=\dfrac{1}{2} \\
\end{align}\]
Now, from equation (i) we get
\[\begin{align}
& \Rightarrow \dfrac{1}{2}+\dfrac{1}{2}+c=4 \\
& \Rightarrow c=3 \\
\end{align}\]
Therefore the \[{{n}^{th}}\] term is given as
\[\Rightarrow {{T}_{n}}=\dfrac{{{n}^{2}}}{2}+\dfrac{n}{2}+3\]
We know that the standard formula of sum of n terms using the \[{{n}^{th}}\] term that is
\[{{S}_{n}}=\sum{{{T}_{n}}}\]
By using the above formula we get the sum of given series as
\[\Rightarrow {{S}_{n}}=\dfrac{1}{2}\left( \sum{{{n}^{2}}}+\sum{n} \right)+3\sum{1}.......equation(iii)\]
We know that the some of the standard results of sum of terms that is
\[\sum{n}=\dfrac{n\left( n+1 \right)}{2}\]
\[\sum{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]
\[\sum{1}=n\]
By using these formulas in equation (iii) we get
\[\Rightarrow {{S}_{n}}=\dfrac{1}{2}\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{n\left( n+1 \right)}{2} \right)+3n\]
Now by taking the common terms out we get
\[\begin{align}
& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( \dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{\left( n+1 \right)}{2} \right)+3 \\
& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( \dfrac{2{{n}^{2}}+3n+1}{6}+\dfrac{\left( n+1 \right)}{2}+6 \right) \\
\end{align}\]
Now, by using the LCM method and adding the terms we get
\[\begin{align}
& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( \dfrac{2{{n}^{2}}+3n+1+3n+3+36}{6} \right) \\
& \Rightarrow {{S}_{n}}=\dfrac{n}{2\times 6}\left( 2{{n}^{2}}+6n+40 \right) \\
& \Rightarrow {{S}_{n}}=\dfrac{n}{6}\left( {{n}^{2}}+3n+20 \right) \\
\end{align}\]
We are given that the sum of series as
\[\Rightarrow {{S}_{n}}=\dfrac{n}{k}\left( {{n}^{2}}+3n+m \right)\]
By comparing this given formula with the result we get
\[\begin{align}
& \Rightarrow m=20 \\
& \Rightarrow k=6 \\
\end{align}\]
Now by finding the value of \[m-k\] we get
\[\Rightarrow m-k=20-6=14\]
Therefore, the value of \[m-k\] is 14.
\[{{S}_{n}}=\sum{{{T}_{n}}}\]
We use some standard results of sum of terms that is
\[\sum{n}=\dfrac{n\left( n+1 \right)}{2}\]
\[\sum{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]
\[\sum{1}=n\]
By using the above results we find the sum of terms to get the required value.
Complete step by step answer:
We are given that the series as \[4+6+9+13+18+..........\]up to \['n'\] terms
Let us assume that the sum of the given series as
\[\Rightarrow {{S}_{n}}=4+6+9+13+.............+{{T}_{n-1}}+{{T}_{n}}...........equation(i)\]
Now let us add ‘0’ on both sides so that the result will not change that is
\[\Rightarrow {{S}_{n}}=0+4+6+9+13+............+{{T}_{n-1}}+{{T}_{n}}...........equation(ii)\]
By subtracting the equation (ii) from equation (i) we get
\[\begin{align}
& \Rightarrow 0=4+\left( 2+3+4+...........\left( n-1 \right)terms \right)-{{T}_{n}} \\
& \Rightarrow {{T}_{n}}-4=2+3+4+...........\left( n-1 \right)terms \\
\end{align}\]
Here we can see that the RHS is in A.P
We know that the sum of \['n'\] terms in an A.P having \['a'\] as first term and \['d'\] as common difference is given as
\[\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\]
By using this formula to above equation we get
\[\begin{align}
& \Rightarrow {{T}_{n}}-4=\dfrac{n-1}{2}\left( 2\times 2+\left( n-1-1 \right)\times 1 \right) \\
& \Rightarrow {{T}_{n}}-4=\dfrac{\left( n-1 \right)\left( n+2 \right)}{2} \\
\end{align}\]
Now, by multiplying the terms on RHS we get
\[\begin{align}
& \Rightarrow {{T}_{n}}=\dfrac{1}{2}\left( {{n}^{2}}+n-2 \right)+4 \\
& \Rightarrow {{T}_{n}}=\dfrac{{{n}^{2}}}{2}+\dfrac{n}{2}+3 \\
\end{align}\]
We know that the standard formula of sum of n terms using the \[{{n}^{th}}\] term that is
\[{{S}_{n}}=\sum{{{T}_{n}}}\]
By using the above formula we get the sum of given series as
\[\Rightarrow {{S}_{n}}=\dfrac{1}{2}\left( \sum{{{n}^{2}}}+\sum{n} \right)+3\sum{1}.......equation(iii)\]
We know that the some of the standard results of sum of terms that is
\[\sum{n}=\dfrac{n\left( n+1 \right)}{2}\]
\[\sum{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]
\[\sum{1}=n\]
By using these formulas in equation (iii) we get
\[\Rightarrow {{S}_{n}}=\dfrac{1}{2}\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{n\left( n+1 \right)}{2} \right)+3n\]
Now by taking the common terms out we get
\[\begin{align}
& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( \dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{\left( n+1 \right)}{2} \right)+3n \\
& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( \dfrac{2{{n}^{2}}+3n+1}{6}+\dfrac{\left( n+1 \right)}{2}+6 \right) \\
\end{align}\]
Now, by using the LCM method and adding the terms we get
\[\begin{align}
& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( \dfrac{2{{n}^{2}}+3n+1+3n+3+36}{6} \right) \\
& \Rightarrow {{S}_{n}}=\dfrac{n}{2\times 6}\left( 2{{n}^{2}}+6n+40 \right) \\
& \Rightarrow {{S}_{n}}=\dfrac{n}{6}\left( {{n}^{2}}+3n+20 \right) \\
\end{align}\]
We are given that the sum of series as
\[\Rightarrow {{S}_{n}}=\dfrac{n}{k}\left( {{n}^{2}}+3n+m \right)\]
By comparing this given formula with the result we get
\[\begin{align}
& \Rightarrow m=20 \\
& \Rightarrow k=6 \\
\end{align}\]
Now by finding the value of \[m-k\] we get
\[\Rightarrow m-k=20-6=14\]
Therefore, the value of \[m-k\] is 14.
Note: We can solve this problem in other methods also.
We have the series as
\[4+6+9+13+18+..........\]
Here, we can see that the differences of consecutive terms are in A.P then the \[{{n}^{th}}\] term is given as
\[\Rightarrow {{T}_{n}}=a{{n}^{2}}+bn+c\]
By taking \[n=1\] we get
\[\Rightarrow a+b+c=4......equation(i)\]
By taking \[n=2\] we get
\[\Rightarrow 4a+2b+c=6......equation(ii)\]
By taking \[n=3\] we get
\[\Rightarrow 9a+3b+c=9......equation(iii)\]
Now by subtracting equation (i) from equation (ii) we get
\[\begin{align}
& \Rightarrow 3a+b=2 \\
& \Rightarrow b=2-3a \\
\end{align}\]
Now, by subtracting equation (ii) from equation (iii) we get
\[\begin{align}
& \Rightarrow 5a+b=3 \\
& \Rightarrow 5a+2-3a=3 \\
& \Rightarrow a=\dfrac{1}{2} \\
\end{align}\]
By substituting the value of \['a'\] in \['b'\] we get
\[\begin{align}
& \Rightarrow b=2-\dfrac{3}{2} \\
& \Rightarrow b=\dfrac{1}{2} \\
\end{align}\]
Now, from equation (i) we get
\[\begin{align}
& \Rightarrow \dfrac{1}{2}+\dfrac{1}{2}+c=4 \\
& \Rightarrow c=3 \\
\end{align}\]
Therefore the \[{{n}^{th}}\] term is given as
\[\Rightarrow {{T}_{n}}=\dfrac{{{n}^{2}}}{2}+\dfrac{n}{2}+3\]
We know that the standard formula of sum of n terms using the \[{{n}^{th}}\] term that is
\[{{S}_{n}}=\sum{{{T}_{n}}}\]
By using the above formula we get the sum of given series as
\[\Rightarrow {{S}_{n}}=\dfrac{1}{2}\left( \sum{{{n}^{2}}}+\sum{n} \right)+3\sum{1}.......equation(iii)\]
We know that the some of the standard results of sum of terms that is
\[\sum{n}=\dfrac{n\left( n+1 \right)}{2}\]
\[\sum{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]
\[\sum{1}=n\]
By using these formulas in equation (iii) we get
\[\Rightarrow {{S}_{n}}=\dfrac{1}{2}\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{n\left( n+1 \right)}{2} \right)+3n\]
Now by taking the common terms out we get
\[\begin{align}
& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( \dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{\left( n+1 \right)}{2} \right)+3 \\
& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( \dfrac{2{{n}^{2}}+3n+1}{6}+\dfrac{\left( n+1 \right)}{2}+6 \right) \\
\end{align}\]
Now, by using the LCM method and adding the terms we get
\[\begin{align}
& \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( \dfrac{2{{n}^{2}}+3n+1+3n+3+36}{6} \right) \\
& \Rightarrow {{S}_{n}}=\dfrac{n}{2\times 6}\left( 2{{n}^{2}}+6n+40 \right) \\
& \Rightarrow {{S}_{n}}=\dfrac{n}{6}\left( {{n}^{2}}+3n+20 \right) \\
\end{align}\]
We are given that the sum of series as
\[\Rightarrow {{S}_{n}}=\dfrac{n}{k}\left( {{n}^{2}}+3n+m \right)\]
By comparing this given formula with the result we get
\[\begin{align}
& \Rightarrow m=20 \\
& \Rightarrow k=6 \\
\end{align}\]
Now by finding the value of \[m-k\] we get
\[\Rightarrow m-k=20-6=14\]
Therefore, the value of \[m-k\] is 14.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
