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Login NowPigeonhole principle states taht if n items are put into m contines with n > m, then at least one container must contain more than one item. For example if one has three gloves then one must have at least two right hand gloves or at least two left hand gloves because one has three objects, but only two categories of handed ness to put then into.

Given recurrance relation is

a_{n} = 3a_{n-1} – 3a_{n-2} + a_{n-3}

Here, degree = 3

The characeristics equation is

r^{3} = 3r^{2} – 3r + 1

or, r^{3} – 3r^{2} – 3r + 1 = 0

Solving, We get

(r-1)(r-1)(r-1) = 0

r = 1 i.e. r_{1} = r_{2} = r_{3} = 1

Now, The general solution of homogenuous equation is:

a_{n} = A_{1}r_{1}^{n} + A_{2}nr_{2}^{n} + A_{3}n^{2}r_{3}^{n}

a_{n} = A_{1}1^{n} + A_{2} . n . (1)^{n} + A_{3} . n^{2} . (1)^{n}

a_{n} = A_{1} + A_{2}n + A_{3}n^{2} ——–(i)

Given condition are

a_{0} = 0 ∴ n = 0

From equation (i):

a_{0} = A_{1} + A_{2} . 0 + A_{3} . 0

∴ A_{1} = 1 ——(ii)

Also, a_{1} = 3 ∴ n = 1

a_{1} = A_{1} + A_{2} + A_{3}

3 = 1 + A_{2} + A_{3}

A_{2} + A_{3} = 2 ——–(iii)

Also, a_{2} = 7 ∴ n = 2

a_{2} = A_{1} + A_{2} . 2 + A_{3} . (2)^{2}

7 = 1 + 2A_{2} + 4A_{3}

2A_{2} + 4A_{3} = 6

A_{2} + 2A_{3} = 3 ——- (iv)

Solving equation (iii) and (iv), We get

A_{3} = 1 and A_{2} = 1

Hence, We get A_{1} = 1, A_{2} = 1 and A_{3} = 1

From equation (i), the general solution is

a_{n} = 1 + n + n^{2}

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