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It isnC0*A^n*b^0 + nC1*A^(n-1)*b^1 + ... + nCr*A^(n-r)*b^r + ... + nCn*A^0*b^n

where nCr = n!/[r!*(n-r)!]

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It isnC0*A^n*b^0 + nC1*A^(n-1)*b^1 + ... + nCr*A^(n-r)*b^r + ... + nCn*A^0*b^n

where nCr = n!/[r!*(n-r)!]

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Any number to the power '0' equals '1'.

Proof ;

Let a^(n) = b

Then dividing

a^(n) / a^(n) = b/b

a^(n-n) = b/b

a^(0) = 1

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The number of links are:

L=b-(n-1)=b-n+1

Where b=Number of branches

n=number of nodes

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Start

dimension A[N], B[N]

For c=1 to N

Input A[N]

Next

For c=1 to N

A[N] = A[N] *10

next

For c+1 to N

B[N] = A[N]

Next

for c=1 to N

print B[N]

Next

End

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The binomial expansion is the expanded form of the algebraic expression of the form (a + b)^n.

There are slightly different versions of Pascal's triangle, but assuming the first row is "1 1", then for positive integer values of n, the expansion of (a+b)^n uses the nth row of Pascals triangle. If the terms in the nth row are p1, p2, p3, ... p(n+1) then the binomial expansion is

p1*a^n + p2*a^(n-1)*b + p3*a^(n-2)*b^2 + ... + pr*a^(n+1-r)*b^(r-1) + ... + pn*a*b^(n-1) + p(n+1)*b^n



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