Its the additive identity prosulate
Answer:
55
Step-by-step explanation:
IHG = 123
IHS = x + 65
SHG = x + 78
IHG = IHS + SHG
123 = x + 65 + x + 78
123 = 2x + 143
123 - 143 = 2x
-20 = 2x
-10 = x
IHS = x + 65
IHS = - 10 + 65
IHS = 55
I'm partial to solving with generating functions. Let

Multiply both sides of the recurrence by
and sum over all
.

Shift the indices and factor out powers of
as needed so that each series starts at the same index and power of
.

Now we can write each series in terms of the generating function
. Pull out the first few terms so that each series starts at the same index
.

Solve for
:

Splitting into partial fractions gives

which we can write as geometric series,


which tells us

# # #
Just to illustrate another method you could consider, you can write the second recurrence in matrix form as

By substitution, you can show that

or

Then solving the recurrence is a matter of diagonalizing the coefficient matrix, raising to the power of
, then multiplying by the column vector containing the initial values. The solution itself would be the entry in the first row of the resulting matrix.
Answer:
40 in. I took the test
Step-by-step explanation: