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Question: Find the sum of \({1^2} - {2^2} + {3^2} - {4^2} + {5^2} - {6^2}...........?\) (Please find out the \...

Find the sum of 12−22+32−42+52−62...........?{1^2} - {2^2} + {3^2} - {4^2} + {5^2} - {6^2}...........? (Please find out the nth{n^{th}} term and then use the sigma method)

Explanation

Solution

Hint : Here in this question first we have to convert the given series into an A.P by solving the successive two terms at a time of the sequence. Then we have to find the nth{n^{th}} term of the A.P
With the help of nth{n^{th}} term we can easily find the sum of our series using the sigma method.
Formula used : an = a1 + (n−1)d{a_n}\, = \,{a_1}\, + \,\left( {n - 1} \right)d
∑0n1=n  and  ∑0nn=n(n+1)2\sum\limits_0^n 1 = n\,\,and\,\,\sum\limits_0^n n = \dfrac{{n\left( {n + 1} \right)}}{2}

Complete step-by-step answer :
In the given question, we have
⇒12−22+32−42+52−62...........\Rightarrow {1^2} - {2^2} + {3^2} - {4^2} + {5^2} - {6^2}...........
We can also write it as
⇒(12−22)+(32−42)+(52−62)...........\Rightarrow \left( {{1^2} - {2^2}} \right) + \left( {{3^2} - {4^2}} \right) + \left( {{5^2} - {6^2}} \right)...........
⇒(1−4)+(9−16)+(25−36)+......\Rightarrow \left( {1 - 4} \right) + \left( {9 - 16} \right) + \left( {25 - 36} \right) + ......
Now, on subtracting
⇒(−3)+(−7)+(−11)+..........\Rightarrow \left( { - 3} \right) + \left( { - 7} \right) + \left( { - 11} \right) + ..........
The above sequence is an Arithmetic Progression with a common difference d = a2−a1d\, = \,{a_2} - {a_1}i.e.
(−7)−(−3)=−4\left( { - 7} \right) - \left( { - 3} \right) = - 4 and the first term is −3 - 3.
Now, let’s find out the nth{n^{th}} term of the given sequence using the formula of nth{n^{th}} term of A.P.
Let an{a_n} be the nth{n^{th}} term of the A.P.
Therefore,
an = a1 + (n−1)d{a_n}\, = \,{a_1}\, + \,\left( {n - 1} \right)d
Here a1=−3  , d=−4{a_1} = - 3\,\,,\,d = - 4
an = −3 + (n−1)−4{a_n}\, = \, - 3\, + \,\left( {n - 1} \right) - 4
Now,
an = −3+4−4n{a_n}\, = \, - 3 + 4 - 4n
an = 1−4n{a_n}\, = \,1 - 4n
So, the nth{n^{th}} term of the given sequence is 1−4n1 - 4n.
Now we will find the sum of sequences using the sigma method.
Sum = ∑0n(1−4n)Sum\, = \,\sum\limits_0^n {\left( {1 - 4n} \right)}
We can also write it as,
Sum = ∑0n1−∑0n4nSum\, = \,\sum\limits_0^n {1 - \sum\limits_0^n {} 4n}
We know that ∑0n1=n  and  ∑0nn=n(n+1)2\sum\limits_0^n 1 = n\,\,and\,\,\sum\limits_0^n n = \dfrac{{n\left( {n + 1} \right)}}{2}
On applying, we get
⇒Sum = n−4∑0nn\Rightarrow Sum\, = \,n - 4\sum\limits_0^n n
⇒Sum = n−4n(n+1)2\Rightarrow Sum\, = \,n - \dfrac{{4n\left( {n + 1} \right)}}{2}
Taking LCM,
⇒(2n−4n2−4n2)\Rightarrow \left( {\dfrac{{2n - 4{n^2} - 4n}}{2}} \right)
⇒(−4n2−2n2)\Rightarrow \left( {\dfrac{{ - 4{n^2} - 2n}}{2}} \right)
On simplification, we get
⇒(−2n2−n)\Rightarrow \left( { - 2{n^2} - n} \right)
⇒−n(2n+1)\Rightarrow - n\left( {2n + 1} \right)
Therefore, the sum of the given sequence is −n(2n+1) - n\left( {2n + 1} \right).
So, the correct answer is “n(2n+1)n\left( {2n + 1} \right)”.

Note : We have solved this question using the sigma method. But there are some other methods also to do this question like we can use the formula of sum of arithmetic progression after converting the given sequence into an A.P. Also, if we want to cross-check the answer by putting the different values of n in the result.