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g(n)=25-49(n-1) Complete the recursive formula of g(n). g(1)= g(n)=g(n-1)+

g(n)=25−49(n−1) g(n)=25-49(n-1) \newlineComplete the recursive formula of g(n) g(n) .\newline g(1) = \(\square\) \newline g(n) = g(n-1)+\(\square\)

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Q. g(n)=25−49(n−1) g(n)=25-49(n-1) \newlineComplete the recursive formula of g(n) g(n) .\newline g(1) = \(\square\) \newline g(n) = g(n-1)+\(\square\)
  1. Find Initial Value: To find the initial value g(1)g(1), we substitute n=1n=1 into the given formula.\newlineg(1)=25−49(1−1)g(1) = 25 - 49(1 - 1)\newlineg(1)=25−49(0)g(1) = 25 - 49(0)\newlineg(1)=25−0g(1) = 25 - 0\newlineg(1)=25g(1) = 25
  2. Find Recursive Formula: Now, we need to find the recursive formula for g(n)g(n). The recursive formula will express g(n)g(n) in terms of g(n−1)g(n-1). To do this, we need to understand how g(n)g(n) changes when we go from n−1n-1 to nn.
    g(n)=25−49(n−1)g(n) = 25 - 49(n - 1)
    g(n−1)=25−49((n−1)−1)g(n-1) = 25 - 49((n-1) - 1)
    g(n−1)=25−49(n−2)g(n-1) = 25 - 49(n - 2)
    Now, let's find the difference between g(n)g(n) and g(n−1)g(n-1).
    g(n)g(n)11
    g(n)g(n)22
    g(n)g(n)33
    g(n)g(n)44
    This means that each term is g(n)g(n)55 less than the previous term.
    So, the recursive formula is:
    g(n)g(n)66

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