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Question: If \({a_n} = {n^2} - 1\) then \({a_{n + 1}}\) is equal to A. \({n^2} + 2n\) B. \({n^2} - 2n\) ...

If an=n2−1{a_n} = {n^2} - 1 then an+1{a_{n + 1}} is equal to
A. n2+2n{n^2} + 2n
B. n2−2n{n^2} - 2n
C. n2+7n{n^2} + 7n
D. n2+9n{n^2} + 9n

Explanation

Solution

According to the question we have to find the value of an+1{a_{n + 1}} when an=n2−1{a_n} = {n^2} - 1. So, first of all we have to replace n with n+1 in the given expression an=n2−1{a_n} = {n^2} - 1. Hence, after solving the obtained expression we can find the value of an+1{a_{n + 1}} as asked in the question.

Formula used: (a+b)2=(a2+b2+2ab)...........(1){(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 an=n2−1{a_n} = {n^2} - 1
⇒an+1=(n+1)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.
⇒an+1=(n2+12+2×n×1)−1 ⇒an+1=n2+2n+1−1 \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.
⇒an+1=n2+2n\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 an+1=n2+2n{a_{n + 1}} = {n^2} + 2n

Therefore option (A) is correct.

Note: If it is given that we have to find an+1{a_{n + 1}} so it is possible only when we replace the n with n+1 in the given expression which is an=n2−1{a_n} = {n^2} - 1
We can also obtain the value of a by placing the value of n if it is given.