Find the sum of the first $50$ terms of the series $a_{n} = -4a_{n-1} + 3$.

Question

I'm not sure where to start this without being given some terms.

Find the sum of the first $50$ terms of the series $$a_{n} = -4a_{n-1} + 3$$ I can see that the common difference is $-4$ and the slope intercept is $3$, but I'm not sure where to go from there.

Answer 1

Consider the first few terms obtained by a direct computation: (assume $a$ is the first term $a_1$) $$a, -4a+3, 16a-9, -64a+39, \cdots, a_n$$ Now, let's write the $n$th partial sum $S_n$: $$S_n=a+(-4a+3)+(16a-9)+(-64a+39)+\cdots+a_n+0$$ Doing a shift using zero we get $$S_n=0+a+(-4a+3)+(16a-9)+\cdots+a_{n-1}+a_n$$ Subtracting term by term: (I don't know how to align these in MathJax, so if someone can please do it, very appreciated) $$0=a+(-5a+3)-4(-5a+3)+16(-5a+3)+\cdots+(-1)^n4^{n-2}-a_n$$ implying $$a_n=a+(\underbrace{1-4+16-\cdots}_{(n-2)\text{ terms}})(-5a+3)$$ You can proceed from here.

Hope this helps. Ask anything if not clear :)

Answer 2

Verify by induction that $$ \displaystyle a_{n} =\frac{3}{5} +( -1)^{n}\left(\frac{3}{5} -a_{1}\right) 4^{n-1}, \displaystyle n\geq 1$$

Then $\sum_{i=1}^{50} a_i=\frac 35\times 50+\Big(\frac 35-a_1\Big)\sum_{i=1}^{50}(-1)^i4^{i-1}$

The second term is a G.P.

Answer 3

Hint

$$a_{n} = -4a_{n-1} + 3$$ Let $a_n=b_n+k$ and replace $$b_n+k=-4b_n-4k+3\implies b_n=-4b_n-(5k-3)$$ So, if you make $k=\frac 35$, you face a geometric progression.

I am sure that you can take it from here.

Answer 4

Note that the sum of the first $n$ terms is: $$\begin{align}a_1&=a_1=\qquad \ \ \ \ \color{blue}{(-4)^0a_1}\\ a_2&=-4a_1+3=-4a_1+3\cdot 1=\color{blue}{(-4)^1a_1}+\color{red}{3\cdot \frac{(-4)^1-1}{-4-1}}\\ a_3&=-4a_2+3=(-4)^2a_1+3(-4+1)=\color{blue}{(-4)^2a_1}+\color{red}{3\cdot\frac{(-4)^2-1}{-4-1}}\\ a_4&=-4a_3+3=(-4)^3a_1+3((-4)^2-4+1)=\color{blue}{(-4)^3a_1}+\color{red}{3\cdot \frac{(-4)^3-1}{-4-1}}\\ \vdots \\ a_n&=-4a_{n-1}+3=(-4)^{n-1}a_1+3((-4)^{n-2}+\cdots+(-4)^2-4+1)=\\ &=\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\color{blue}{(-4)^{n-1}a_1}+\color{red}{3\cdot \frac{(-4)^{n-1}-1}{-4-1}} \\ \hline S_n&=\color{blue}{a_1[1-4+(-4)^2+\cdots+(-4)^{n-1}]}+\\ &+\color{red}{3\cdot \left[\frac{(-4)^1-1}{-4-1}+\frac{(-4)^2-1}{-4-1}+\cdots +\frac{(-4)^{n-1}-1}{-4-1}\right]}=\\ &=a_1\cdot \frac{(-4)^n-1}{-4-1}+3\cdot \frac{\frac{(-4)((-4)^{n-1}-1)}{-4-1}-(n-1)}{-5}=\\ &=\frac1{25} \left[(3-5 a_1) (-4)^n + 5 a_1 + 15 n - 3\right]. \end{align} $$