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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 \[m-k\]

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Last updated date: 13th Jun 2024
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Answer
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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.