Method of Difference

Method of Differences

The method of differences is a method to solve the sequence and series questions that can not be solved using the concepts of Arithmetic Progression and Geometric Progression. In this method, we try to get the AP or GP series by subtracting the subsequent terms of the series.

Steps to solve questions using the Method of Differences

Following is the step-by-step procedure to calculate a series by using the method of differences :

Step 1: Suppose $a_1, a_2, a_3, \ldots \ldots$ is a sequence such that the sequence is neither an Arithmetic Progression nor a Geometric progression.

Step 2: Find the difference between subsequent terms of the series. We do not subtract the first and last term of the series.

Step 3: Check if $a_2-a_1, a_3-a_3, \ldots$ is either an. A.P. or a G.P.
Step 4: When we successfully determine the nature of the series, we calculate the sum of the series for ( $\mathrm{n}-1$ ) terms.

Step 5: Now, we have a simplified solution of the series.

Method of Differences (Shortcut)

To find the $\mathrm{n}^{\text {th }}$ term of the series, you can use the following steps:
1. If the sequence of the first consecutive difference is in A.P., then the $n^{\text {th }}$ term, $T_n=a n^2+b n+c$ or $a(n-1)(n-2)+b(n-1)+$ $c$, where $\mathrm{a}, \mathrm{b}$, and c are constants. To find the value of $a, b$, and $c$ put $n=1,2,3$ and put the value of $T_1, T_2, T_3$.
2. If the sequence of the first consecutive difference is in G.P., then the $n$th term, $T_n=a r^n+b$, where $a$ and $b$ are constant terms and $r$ is the common ratio of GP. To find the value of a and $b$ put $n=1,2$ and put the value of $T_1, T_2$

Recommended Video Based on Method of Differences:

Solved Examples Based on the Method of Differences

Example 1: Suppose f is a function satisfying $\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y})$ for all $x, y \in \mathbb{N}$ and $f(1)=\frac{1}{5} \sum_{\text {If }}^{m=1} \frac{f(n)}{n(n+1)(n+2)}=\frac{1}{12}$, then $m$ is equal to
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Solution

$
\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y}) \forall \mathrm{x}, \mathrm{y} \in \mathrm{N} \text { and } \mathrm{f}(1)=\frac{1}{5}
$

for $x=y=1$

$
\begin{aligned}
& \mathrm{f}(2)=\mathrm{f}(1)+\mathrm{f}(1)=2 \mathrm{f}(1) \\
& \mathrm{f}(3)=\mathrm{f}(2+1)=\mathrm{f}(2)+\mathrm{f}(1)=3 \mathrm{f}(1)
\end{aligned}
$

in General

$
f(n)=n f(1)=\frac{n}{5}
$
$\begin{aligned}
& \sum_{\mathrm{n}=1}^{\mathrm{m}} \frac{\mathrm{f}(\mathrm{n})}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}=\frac{1}{12} \\
& \Rightarrow \sum_{\mathrm{n}=1}^{\mathrm{m}} \frac{\mathrm{n}}{5 \mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}=\frac{1}{12} \\
& \Rightarrow \frac{1}{5} \sum_{\mathrm{n}=1}^{\mathrm{m}} \frac{1}{(n+1)(n+2)}=\frac{1}{12} \\
& \Rightarrow \sum_{\mathrm{n}=1}^{\mathrm{m}}\left(\frac{1}{\mathrm{n}+1}-\frac{1}{\mathrm{n}+2}\right)=\frac{5}{12} \\
& \Rightarrow\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots \ldots .+\left(\frac{1}{m+1}-\frac{1}{m+2}\right)=\frac{5}{12} \\
& \Rightarrow \frac{1}{2}-\frac{1}{m+2}=\frac{5}{12} \\
& \Rightarrow \frac{1}{\mathrm{~m}+2}=\frac{1}{2}-\frac{5}{12}=\frac{1}{12} \\
& \Rightarrow \mathrm{m}=10
\end{aligned}$

Hence, the answer is the 10.

Example 2: Find the sum of series $1+3+7+13+\ldots+211$

Solution

$
S_n=1+3+7+13+21 \ldots
$
Differences are $2,4,6,8, \ldots$ which is an $A P$
For the difference in A.P., By the above concept

$
T_n=a n^2+b n+c
$

put $n=1,2$, and 3

$
T_n=n^2-n+1
$
For last term 211, $n=15$

$
\begin{aligned}
& S_n=\sum_1^{15} T_n \\
& S_n=\frac{n(n+1)(2 n+1)}{6}-\frac{n(n+1)}{2}+n \\
& S_n=\frac{1}{3}\left(n^3+2 n\right) \\
& S_{15}=1135
\end{aligned}
$

Hence, the answer is 1135

Example 3: Find the sum first 8 terms of the series

$
1+3+7+15+29 \ldots
$

Solution

Now, the given series is

$
1+3+7+15+29 \ldots
$

difference of consecutive terms $=2+4+8+14 \ldots$
difference of this difference $=2+4+6+.$.
This is in A.P.
by above concept

$
T_n=a n^3+b n^2+c n+d
$

Put $n=1,2,3,4$ and solving the equations formed, we get

$
\begin{aligned}
& d=-1, c=8 / 3, b=-1, a=\frac{1}{3} \\
& T_n=\frac{1}{3}\left[n^3-3 n^2+8 n-3\right] \\
& S_n=\sum_1^8 T_n \\
& S_8=148
\end{aligned}
$

Hence, the answer is 148

Example 4: The $10^{th}$ term of the series $2+8+20+40+70+112+\ldots \ldots$.

Solution
The sequence of first consecutive differences is $6,12,20,30,42$.
The sequence of second consecutive differences is $6,8,10,12 \ldots \ldots$ it is an AP

$
\begin{aligned}
& S_n=2+8+20+40+70+112+\ldots \ldots+T_{n-1}+T_n \\
& S_n=2+8+20+40+70+112+\ldots \ldots+T_{n-1}+T_n \\
& 0=2+6+12+20+30+42+\ldots \ldots+\left(T_{n-1}-T_n\right)-T_n \\
& T_n=2+6+12+20+30+42+\ldots \ldots \ldots+\text { upto } n \text { term }
\end{aligned}
$

Or

$
\begin{aligned}
& T_n=2+6+12+20+30+42+\ldots \ldots \ldots+t_{n-1}+t_n \\
& T_n=2+6+12+20+30+42+\ldots \ldots \ldots+t_{n-1}+t_n
\end{aligned}
$

Substract (iii) - (iv)

Hence, the answer is 440

$
\begin{aligned}
& t_n=2+4+6+8+10+12+\ldots \ldots \ldots \ldots\left(t_{n-1}-t_n\right) \\
& t_n=\frac{n}{2}(2 \times 2+(n-1) \times 2)=n^2+n
\end{aligned}
$
Example 5: The sum of 8 terms of series $1+5+13+29 \ldots$. is:

Solution: Differences in consecutive terms are $4,8,16$ $\qquad$ which is a GP with $r=2$

So, let $a_n=a .2^n+c$
Put $\mathrm{n}=1: a_1=a .2+c \Rightarrow 2 a+c=1$
Put $\mathrm{n}=2: a_2=a .2^2+c \Rightarrow 4 a+c=5$
(2) - (1) $: 2 a=4 \Rightarrow a=2$

Using (1) $\Rightarrow c=-3$
So, $a_n=2.2^n-3$

$
\begin{aligned}
\sum a_n & =2 \sum 2^n-3 \sum 1 \\
& =2\left(2+2^2+2^3+\ldots .2^n\right)-3 . n \\
& =2.2 \frac{\left(2^n-1\right)}{(2-1)}-3 n \\
& =4\left(2^n-1\right)-3 n
\end{aligned}
$

So, $S-8=4 .\left(2^8-1\right)-24=996$
Hence, the answer is 996 .