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Question:

What is the sum up to $n$ terms of the following series? $$3+8+22+72+266+1036\dots$$

My Approach: $$S_n=3+8+22+72+266+1036\dots T_n$$ $$-S_n=0-3-8-22-72-266-1036\dots-T_n$$ $$T_n=3+5+14+50+194+770\dots ({T_n}-{T_{n-1}})$$ $$-T_n=0-3-5-14-50-194-770\dots-({T_n}-{T_{n-1}})$$ $$({T_n}-{T_{n-1}})=3+2+9+36+144+576\dots+({T_{n-1}}-{T_{n-2}})$$ $$({T_n}-{T_{n-1}})=5+9+36+144+576\dots+({T_{n-1}}-{T_{n-2}})$$ Now, this forms a $GP$ with $a=9$ and $r=4$. $$\therefore ({T_n}-{T_{n-1}})=5+\sum^{n-2}_{r=1}9.4^{r-1}$$ $$\implies ({T_n}-{T_{n-1}})=5+3(4^{n-2}-1)$$ $$\implies ({T_n}-{T_{n-1}})=2+3.4^{n-2}$$ $${T_n}={T_{n-1}}+2+3.4^{n-2}$$ Now, $$T_n=3+5+14+50+194+770\dots ({T_n}-{T_{n-1}})$$ $$\implies T_n=3+5+14+50+194+770\dots ({T_{n-1}}+2+3.4^{n-2}-{T_{n-1}})$$ $$T_n=3+5+14+50+194+770\dots 2+3.4^{n-2}$$

Which seems wrong to me, and I have no idea how to proceed further. Please help.

General Kenobi
  • 439

2 Answers2

1

Let $a_0=3,a_1=8,a_2= 22, \ldots$ etc.

So, you are looking for a term describing $\sum_{k=0}^na_k$.

The second differences of the members of the given sequence $(a_k)$ are $e_k=9\cdot 4^k$ ($k=0,1,2,\ldots$).

Hence, the first differences are $$d_n = 5 +9\sum_{k=0}^{n-1}4^k=2+3\cdot 4^n$$

This gives

$$a_n = 3+\sum_{k=0}^{n-1}(2+3\cdot 4^k)=2(n+1)+4^n$$

Finally, summing $a_k$ gives

$$s_n= \sum_{k=0}^na_k = \sum_{k=0}^n(2(k+1)+4^k)$$ $$=\boxed{(n+1)(n+2)+\frac 13(4^{n+1}-1)},\; n=0,1,2,\ldots$$

trancelocation
  • 32,243
1

The sequence: $$3,8,22.72,266,1036,....$$ The first diferences are: $$5, 14,50,194,760,.....$$ Second differences are in GP with common ratio 0f $4$: $$ 9, 36, 144, 576,....$$ So the $k$ term is $T(k)=An+B+C4^k$ So we have $$T(1)=A+B+4C=3, T(2)=2A+B+16C=8, T(3)=3A+B+64C=22$$ Solving these eqns. we Finally $$S(n)=\sum_{k=1}^{n} T(k)=2 \sum_{k=1}^n k+\sum_{k=1}^{n} 4^{k-1}=n(n+1)+\frac{4^n-1}{3 }$$

Z Ahmed
  • 43,235