What is the sum of the last five coefficients in the expansion of (1 + x)⁹ when it is expanded in ascending powers of x?

Concept:

(1 + x)ⁿ  = ⁿΣ_r-0 nC_r . x^r = [C₀ + C₁ x + C₂ x² + … C_n x_n]

\(\large{\rm n_{C_{n-r}}=n_{C_{r}}}\)

Calculation:

\(\begin{aligned} &(1+\mathrm{x})^{9}={ }^{9} \mathrm{C}_{0}+{ }^{9} \mathrm{C}_{1} \mathrm{x}+{ }^{9} \mathrm{C}_{2} \mathrm{x}^{2}+\ldots . .^{9} \mathrm{C}_{9} \mathrm{x}^{9}\\ &\text { Substituting } \mathrm{x}=1, \text { we get }\\ &2^{9}={ }^{9} \mathrm{C}_{0}+{ }^{9} \mathrm{C}_{1}+{ }^{9} \mathrm{C}_{2}+\ldots .^{9} \mathrm{C}_{9}----(1)\end{aligned} \)

As we know, ⁿC_r = ⁿC_n-r

So,

\({ }^{9} \mathrm{C}_{0}+{ }^{9} \mathrm{C}_{1}+ +{ }^{9} \mathrm{C}_{2} +{ }^{9} \mathrm{C}_{3}+ ^{9} \mathrm{C}_{4} ={ }^{9} \mathrm{C}_{9}+{ }^{9} \mathrm{C}_{8}+{ }^{9} \mathrm{C}_{7}+{ }^{9} \mathrm{C}_{6}+{ }^{9} \mathrm{C}_{5} \)     ----(2)

Now, From (1) and (2), we get,

\(2^{9}=2\left[{ }^{9} \mathrm{C}_{9}+{ }^{9} \mathrm{C}_{8}+{ }^{9} \mathrm{C}_{7}+9 \mathrm{C}_{6}+{ }^{9} \mathrm{C}_{5}\right]\)

⇒ \(\left[{ }^{9} \mathrm{C}_{9}+{ }^{9} \mathrm{C}_{8}+{ }^{9} \mathrm{C}_{7}+9 \mathrm{C}_{6}+{ }^{9} \mathrm{C}_{5}\right] = 2^{8}\)

⇒ 256

Hence, option (1) is correct.