Find the coefficient in front of the ninth power of x

Given: \((1+x)^9 + (1+x)^{10} + (1+x)^{11} + (1+x)^{12} + (1+x)^{13} + (1+x)^{14}\)

Asked: Find the coefficient in front of the ninth power of x (\(x^9\)) when all the brackets are expanded.

Coefficient of x^m in the expansion of (1+x)^n = nCm = nC(n-m)

Let us find coefficient in front of the ninth power of x (\(x^9\)) when all the brackets are expanded for each term separately

\((1+x)^9\)
Coefficient of \(x^9\) = 1

\((1+x)^{10}\)
Coefficient of \(x^9\) = 10C1 = 10

\((1+x)^{11}\)
Coefficient of \(x^9\) = 11C2= 55

\((1+x)^{12}\)
Coefficient of \(x^9 \)= 12C3= 220

\((1+x)^{13}\)
Coefficient of \(x^9 \)= 13C4 = 715

\((1+x)^{14}\)
Coefficient of \(x^9\) = 14C5 = 2002

Coefficient of \(x^9\) in the expansion of \((1+x)^9 + (1+x)^{10} + (1+x)^{11} + (1+x)^{12} + (1+x)^{13} + (1+x)^{14}\) = 1 + 10 + 55 + 220 + 715 + 2002 = 3003

IMO A

Can be found with binomial formula for the kth term in the expansion
\((X+Y)^N = (N)Ck * X^k * Y ^{N-k} \\
\)
coefficient of x^9 in (1+x)^9 is 9C9
coefficient of x^9 in (1+x)^10 is 10C9

And so on

the coefficient in front of the ninth power of x (x9x9) when all the brackets are expanded

= 9C9+10C9+11C9+12C9+13C9+14C9

Yes .. Correct.. Thank you for pointing .

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