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# What will the nth term of the series 3+7+13+21+…..?${\text{A}}{\text{. }}4n - 1 \\ {\text{B}}{\text{. }}{n^n} + 2n \\ {\text{C}}{\text{. }}{n^2} + n + 1 \\ {\text{D}}{\text{. }}{n^2} + 2 \\$

Last updated date: 23rd Mar 2023
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Hint: Here, we will proceed by considering all the nth terms given in the options and then will put n=1,2,3,.. in order to find the reassemble of the obtained values of the terms with those of the given series.

The given sum of series is 3+7+13+21+…..
Let us consider all the nth terms given by obtains mentioned in the problem.
Consider the nth term as ${a_n} = 4n - 1$.
Put n=1 for the first term of the series whose nth term is given by ${a_n} = 4n - 1$, we get
${a_1} = \left( {4 \times 1} \right) - 1 = 4 - 1 = 3$
Put n=2 for the second term of the series whose nth term is given by ${a_n} = 4n - 1$, we get
${a_2} = \left( {4 \times 2} \right) - 1 = 8 - 1 = 7$
Put n=3 for the third term of the series whose nth term is given by ${a_n} = 4n - 1$, we get
${a_3} = \left( {4 \times 3} \right) - 1 = 12 - 1 = 11$
But, since the third term of the given series is 13 which is not equal to 11 i.e., the third term of the series whose nth term is given by ${a_n} = 4n - 1$.
So, option A is incorrect.
Consider the nth term as ${a_n} = {n^n} + 2n$.
Put n=1 for the first term of the series whose nth term is given by ${a_n} = {n^n} + 2n$, we get
${a_1} = {\left( 1 \right)^1} + \left( {2 \times 1} \right) = 1 + 2 = 3$
Put n=2 for the second term of the series whose nth term is given by ${a_n} = {n^n} + 2n$, we get
${a_2} = {\left( 2 \right)^2} + \left( {2 \times 2} \right) = 4 + 4 = 8$
But, since the second term of the given series is 7 which is not equal to 8 i.e., the second term of the series whose nth term is given by ${a_n} = {n^n} + 2n$.
So, option B is also incorrect.
Consider the nth term as ${a_n} = {n^2} + n + 1$.
Put n=1 for the first term of the series whose nth term is given by ${a_n} = {n^2} + n + 1$, we get
${a_1} = {\left( 1 \right)^2} + 1 + 1 = 1 + 2 = 3$
Put n=2 for the second term of the series whose nth term is given by ${a_n} = {n^2} + n + 1$, we get
${a_2} = {\left( 2 \right)^2} + 2 + 1 = 4 + 3 = 7$
Put n=3 for the third term of the series whose nth term is given by ${a_n} = {n^2} + n + 1$, we get
${a_3} = {\left( 3 \right)^2} + 3 + 1 = 9 + 4 = 13$
Here, these values are the same as those of the given series. Hence, the nth term of the given series is given by ${a_n} = {n^2} + n + 1$.
So, option C is correct.
Consider the nth term as ${a_n} = {n^2} + 2$.
Put n=1 for the first term of the series whose nth term is given by ${a_n} = {n^2} + 2$, we get
${a_1} = {\left( 1 \right)^2} + 2 = 1 + 2 = 3$
Put n=2 for the second term of the series whose nth term is given by ${a_n} = {n^2} + 2$, we get
${a_2} = {\left( 2 \right)^2} + 2 = 4 + 2 = 6$
But, since the second term of the given series is 7 which is not equal to 6 i.e., the second term of the series whose nth term is given by ${a_n} = {n^2} + 2$.
So, option D is incorrect.

Note: In these types of problems, we eliminate the wrong options by putting different values of n starting from 1 in order to obtain first, second and so on up to that value of n which comes out to be different from those in the given series.