Question:

<p><strong>Solution:</strong></p><p>Given: <span class="math">$(1 + x^2)^{40} \times \left(x^{-5/2} + 2 + \frac{2}{x^{1/2}}\right)$</span></p><p>Rewrite the expression:</p><p><span class="math">$$= (1 + x^2)^{40} \times \left(x + \frac{1}{x}\right)^2$$</span></p><p><span class="math">$$= (1 + x^2)^{40} \times \left(x + \frac{1}{x}\right)^2$$</span></p><p><span class="math">$$= x^{10}(1 + x^2)^{40}(1 + x^2)^{-10}$$</span></p><p><span class="math">$$= x^{10}(1 + x^2)^{30}$$</span></p><p>To find the coefficient of <span class="math">$x^{20}$</span>, we need the coefficient of <span class="math">$x^{10}$</span> in <span class="math">$(1 + x^2)^{30}$</span>.</p><p>In the binomial expansion of <span class="math">$(1 + x^2)^{30}$</span>, the general term is:</p><p><span class="math">$$^{30}C_r (x^2)^r = ^{30}C_r x^{2r}$$</span></p><p>For <span class="math">$x^{10}$</span>, we need <span class="math">$2r = 10$</span>, so <span class="math">$r = 5$</span>.</p><p>Therefore, the coefficient is <span class="math">$^{30}C_5 = ^{30}C_{25}$</span>.</p><p>∴ Answer is (b).</p>