Answer:
Tn = Tn-1 + 2(n-1) + 5
Kindly note that Tn-1 means T subscript n-1
Step-by-step explanation:
Here, we want an expression for the nth term.
First term is 7
Then first common difference is 7
second common difference is 7 + 2
Third common difference is 9 + 2
So within the common differences, the nth term is 7 + (n-1) 2
Now, the nth term of the series would be;
Tn = Tn-1 + 7 + 2(n-1)
Tn = Tn-1 + 7 + 2n -2
Tn = Tn-1 + 2n + 5
Now there is a fix to this,
n for the term is not the same n for the common difference.
the 7th term works with the 6th common. difference, while the 8th term work for the 7th common difference.
So we might need to rewrite the final expression as follows;
Tn = Tn-1 + 2(n-1) + 5
We must find UNIQUE combinations because choosing a,b,c,d... is the same as d,c,b,a...etc. For this type of problem you use the "n choose k" formula...
n!/(k!(n-k)!), n=total number of choices available, k=number of choices made..
In this case:
20!/(10!(20-10)!)
20!/(10!*10!)
184756
-0.8 = -0.8/10
-0.8 x 10 = -8
<span>1 × 10 = 10</span>
<span>-8/10
Simplified = -4/5
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Hi there, we use PEMDAS to solve this problem, PEMDAS is parenthesis, Exponents, Multiply, Divide, Add, and Subtract. In this case, we have [6.6÷(-5+3)]*(-1), 6.6/-5+3(-1)=6.6/-2(-1)=(-3.3)(-1)=3.3. So, your answer is 3.3
