Question

(3) 3
(4)
3
10.
Find the nth term of the series
3 + 7 + 13 + 21+ ....
(A) n2 + 1
(2) n+ n + 1
(C)n(n+1)

(4) n° + n + 1

Answer 1

Answer:

We will subtract 1 from the sum of (N+1) terms to get the answer. Terms are : 1, 3, 7, 13, 21, 31, 43, 57, ...........

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Answer 2

Answer:

Two ways you can solve this problem are visually and mathematically. Both will give you the final answer of 57.

Visual Method

Between each of the numbers, the difference is increasing by 2 each time. 3 comes 2 after 1, 7 comes 4 after 3, 13 comes 6 after 7, and so on. Since 43 comes 12 after 31, the next number would come 14 after 43. Because of this, 43 + 14 is 57 –– your answer.

Mathematical Method

Say you want to calculate the nth number of the sequence. It would be impractical to keep writing the numbers, because as you approach infinity, the numbers will get larger and larger. I will solve this problem recursively, which you can read more about here - .

Let f(n) be the function that calculates the nth term of the sequence. Take, for example, the third term of the sequence, which is 7 -- this gives us n=3 . In order to get to the number 7 , we had to add the previous number in the sequence, or f(n−1) , with a certain even number, which is 2(n−1) .

Just so you can check my conclusion, the previous number in the sequence was 3 . We added 2(3−1) or 4 to that number, which gave us 7 . Therefore, the sequence can be represented by this function:

f(n)=f(n−1)+2(n−1)

I implemented this recursive function in Python. Here it is:

def sequence(num):

if (num == 1):

return 1

else:

return sequence(num - 1) + 2 * (num - 1)

Hope this helps! Please mark the answer as the branliest and verified.