Let a_n_n=1^∞ be a sequence such that a_1=1,a_2=1 and a_n+2=2a_n+1+a_n for all n≥ 1. Then the value of 47Σ _n=1^∞ ( a_n 2^3n) is equal to .

We have,

$$47\sum _{n=1}^{\infty }(\frac{{a}_{n}}{{2}^{3n}})$$47n=1∑∞​(23nan​​)

$\displaystyle \Rightarrow \sum _{n=1}^{\infty }(\frac{{a}_{n}}{{8}^{n}})=P$⇒n=1∑∞​(8nan​​)=P(let)

And,

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

$$\Rightarrow \frac{{a}_{n+2}}{{8}^{n}}=\frac{2{a}_{n+1}}{{8}^{n}}+\frac{{a}_{n}}{{8}^{n}}$$⇒8nan+2​​=8n2an+1​​+8nan​​

$$\Rightarrow 64(\frac{{a}_{n+2}}{{8}^{n+2}})=\frac{16{a}_{n+1}}{{8}^{n+1}}+\frac{{a}_{n}}{{8}^{n}}$$⇒64(8n+2an+2​​)=8n+116an+1​​+8nan​​

$$\Rightarrow 64\sum _{n=1}^{\infty }(\frac{{a}_{n+2}}{{8}^{n+2}})=16\sum _{n=1}^{\infty }(\frac{{a}_{n+1}}{{8}^{n+1}})+\sum _{n=1}^{\infty }(\frac{{a}_{n}}{{8}^{n}})$$⇒64n=1∑∞​(8n+2an+2​​)=16n=1∑∞​(8n+1an+1​​)+n=1∑∞​(8nan​​)

$$\Rightarrow 64(P-\frac{{a}_{1}}{8}-\frac{{a}_{2}}{{8}^{2}})=16(P-\frac{{a}_{1}}{8})+P$$⇒64(P−8a1​​−82a2​​)=16(P−8a1​​)+P

$$\Rightarrow 64(P-\frac{1}{8}-\frac{1}{64})=16(P-\frac{1}{8})+P$$⇒64(P−81​−641​)=16(P−81​)+P

$$\Rightarrow 64P-8-1=16P-2+P$$⇒64P−8−1=16P−2+P

$$\Rightarrow 47P=7$$⇒47P=7

$$\Rightarrow 47\sum _{n=1}^{\infty }(\frac{{a}_{n}}{{8}^{n}})=7$$⇒47n=1∑∞​(8nan​​)=7