Question

If ${a_n} = {n^2} - 1$ then ${a_{n + 1}}$ is equal to
A. ${n^2} + 2n$
B. ${n^2} - 2n$
C. ${n^2} + 7n$
D. ${n^2} + 9n$

Answer 1

Hint: According to the question we have to find the value of ${a_{n + 1}}$ when ${a_n} = {n^2} - 1$. So, first of all we have to replace n with n+1 in the given expression ${a_n} = {n^2} - 1$. Hence, after solving the obtained expression we can find the value of ${a_{n + 1}}$ as asked in the question.

Formula used: ${(a + b)^2} = ({a^2} + {b^2} + 2ab)...........(1)$

Complete step-by-step answer:
Step 1: so first of all we have to replace the term n with n+1 in the given expression ${a_n} = {n^2} - 1$
$ \Rightarrow {a_{n + 1}} = {(n + 1)^2} - 1$
Step 2: Now, we have to solve the obtained expression as in the step 1 we have to use the formula (1) as mentioned in the solution hint.
$
   \Rightarrow {a_{n + 1}} = ({n^2} + {1^2} + 2 \times n \times 1) - 1 \\
   \Rightarrow {a_{n + 1}} = {n^2} + 2n + 1 - 1
 $
On solving the expression obtained just above.
$ \Rightarrow {a_{n + 1}} = {n^2} + 2n$
Final solution: hence, with the help of the formula (1) and replacing n with n+1 we have obtained the value of ${a_{n + 1}} = {n^2} + 2n$

Therefore option (A) is correct.

Note: If it is given that we have to find ${a_{n + 1}}$ so it is possible only when we replace the n with n+1 in the given expression which is ${a_n} = {n^2} - 1$
We can also obtain the value of a by placing the value of n if it is given.