ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2559 (วิชาสามัญเก่า)

​$$\begin{gathered} \mathrm{จากโจทย์} 𝑓\left(\mathrm{x}\right)=\sum _{\mathrm{k}=1}^{100}\mathrm{k}·{\mathrm{x}}^{2\mathrm{k}-1} \\ \frac{1}{\sqrt{2}}𝑓\left(\mathrm{x}\right)=\frac{1}{\sqrt{2}}\sum _{\mathrm{k}=1}^{100}\mathrm{k}·{\mathrm{x}}^{2\mathrm{k}-1} \\ \mathrm{แทน} \mathrm{x}=\sqrt{2} \frac{1}{\sqrt{2}}𝑓\left(\sqrt{2}\right)=\frac{1}{\sqrt{2}}\sum _{\mathrm{k}=1}^{100}\mathrm{k}·{\sqrt{2}}^{2\mathrm{k}-1} \\ =\frac{1}{\sqrt{2}}\sum _{\mathrm{k}=1}^{100}\frac{\mathrm{k}·{\sqrt{2}}^{2\mathrm{k}}}{\sqrt{2}} \\ =\frac{1}{2}\sum _{\mathrm{k}=1}^{100}\mathrm{k}·2^{\mathrm{k}} \\ =\frac{1}{2}\left(1·2^1+2·2^2+3·2^3+...+100·2^{100}\right) \\ =1·2^0+2·2^1+3·2^2+...+100·2^{99} \end{gathered}$$

$$\begin{gathered} \mathrm{กำหนดให้} \mathrm{S}=\frac{1}{\sqrt{2}}𝑓\left(\sqrt{2}\right) \\ \mathrm{จะได้} \mathrm{S}=1·2^0+2·2^1+3·2^2+...+100·2^{99}\to \boxed{1} \\ \mathrm{นำ} \boxed{1}\times 2 2\mathrm{S}=1·2^1+2·2^2+...+99·2^{99}+100·2^{100}\to \boxed{2} \\ \mathrm{นำ} \boxed{1}-\boxed{2} -\mathrm{S}=1·2^0+1·2^1+1·2^2+...+1·2^{99}-100·2^{100} \\ -S=1·2^0+2^1+2^2+...+2^{99}-100·2^{100} \\ -S=(1+2^1+2^2+...+2^{99})-100·2^{100}\to \boxed{3} \end{gathered}$$

$$\begin{gathered} \mathrm{สูตรอนุกรมมเรขาคณิต} {\mathrm{S}}_{\mathrm{n}}=\frac{{\mathrm{a}}_1\left({\mathrm{r}}^{\mathrm{n}}-1\right)}{\mathrm{r}-1} \\ \mathrm{โดย} {\mathrm{a}}_1=1 , \mathrm{r}=2 , \mathrm{n}=100 \\ \mathrm{จาก} \boxed{3} \mathrm{จะได้} -\mathrm{S}=\frac{1\left(2^{100}-1\right)}{2-1}-100·2^{100} \\ -\mathrm{S}=2^{100}-1-100·2^{100} \\ -\mathrm{S}=-1-99·2^{100} \\ \mathrm{S}=1+99·2^{100} \\ \mathrm{ดังนั้น} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \frac{1}{\sqrt{2}}𝑓\left(\sqrt{2}\right)=1+99·2^{100} \end{gathered}$$