Solutions:

\( a_{n+2} - a_{n+1} - a_n + 1 \)

\( n = 0 \quad a_2 - a_1 = a_1 - a_0 + 1 \)

\( n = 1 \quad a_3 - a_2 = a_2 - a_1 + 1 \)

\( n = 2 \quad a_4 - a_3 = a_3 - a_2 + 1 \)

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\( a_{n+2} - a_{n+1} = a_{n+1} - a_n + 1 \)

Adding \( \ a_{n+2} - a_1 = a_{n+1} - a_0 + (n+1) \)

\( \Rightarrow \ a_{n+2} - a_{n+1} = n+1 \)

\( n = 0 \quad a_2 - a_1 = 1 \)

\( n = 1 \quad a_3 - a_2 = 2 \)

\( n = 2 \quad a_4 - a_3 = 3 \)

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\( a_{n+2} - a_{n+1} = n+1 \Rightarrow a_{n+2} = \dfrac{(n+1)(n+2)}{2} \)

\( \Rightarrow \ a_n = \dfrac{(n-1)n}{2} \Rightarrow \dfrac{1}{2} \sum_{n=2}^{\infty} \dfrac{(n-1)n}{7^n} \)

\( \dfrac{1}{2} \left( \dfrac{1 \cdot 2}{7^2} + \dfrac{2 \cdot 3}{7^3} + \dfrac{3 \cdot 4}{7^4} + \ldots \right) \)

\( S_{\infty} = \dfrac{1 \cdot 2}{7^2} + \dfrac{2 \cdot 3}{7^3} + \dfrac{3 \cdot 4}{7^4} + \dfrac{4 \cdot 5}{7^5} + \ldots \)

\( \dfrac{1}{7} S_n = \dfrac{1 \cdot 2}{7^3} + \dfrac{2 \cdot 3}{7^4} + \dfrac{3 \cdot 4}{7^5} + \ldots \)

JEE MainsJEE AdvanceBITSATJEE Mains | PYQ

JEE Mains JEE Advance BITSAT
JEE Mains | PYQ