JEE Main 2022 — Mathematics Algebra

Given ${a}_{n+2}=2{a}_{n+1}-{a}_{n}+1$an+2​=2an+1​−an​+1

Dividing by ${7}^{n+2}$7n+2, we get

$$\frac{{a}_{n+2}}{{7}^{n+2}}=\frac{2}{7}\cdot \frac{{a}_{n+1}}{{7}^{n+1}}-\frac{1}{49}\cdot \frac{{a}_{n}}{{7}^{n}}+\frac{1}{{7}^{n+2}}$$7n+2an+2​​=72​⋅7n+1an+1​​−491​⋅7nan​​+7n+21​

So, $\displaystyle \sum _{n=2}^{\infty }\frac{{a}_{n+2}}{{7}^{n+2}}=\frac{2}{7}\sum _{n=2}^{\infty }\frac{{a}_{n+1}}{{7}^{n+1}}-\frac{1}{49}\sum _{n=2}^{\infty }\frac{{a}_{n}}{{7}^{n}}+\sum _{n=2}^{\infty }\frac{1}{{7}^{n+2}}$n=2∑∞​7n+2an+2​​=72​n=2∑∞​7n+1an+1​​−491​n=2∑∞​7nan​​+n=2∑∞​7n+21​

Let

$$\sum _{n=2}^{\infty }\frac{{a}_{n}}{{7}^{n}}=P$$n=2∑∞​7nan​​=P

$$\Rightarrow (P-\frac{{a}_{3}}{{7}^{3}}-\frac{{a}_{2}}{{7}^{2}})=\frac{2}{7}(P-\frac{{a}_{2}}{{7}^{2}})-\frac{1}{49}P+\frac{1}{{7}^{4}}(\frac{1}{1-\frac{1}{7}})$$⇒(P−73a3​​−72a2​​)=72​(P−72a2​​)−491​P+741​(1−71​1​)

(\because {a}_{2}=1&{a}_{3}=3)

$$\Rightarrow P(1-\frac{2}{7}+\frac{1}{49})=\frac{3}{{7}^{3}}+\frac{1}{{7}^{2}}-\frac{2}{{7}^{3}}+\frac{1}{6\cdot {7}^{3}}$$⇒P(1−72​+491​)=733​+721​−732​+6⋅731​

$\Rightarrow \frac{36P}{49}=\frac{1}{42}$⇒4936P​=421​ $\Rightarrow P=\frac{7}{216}$⇒P=2167​