Re-investment of interest Thomas invests X into Fund $1$ at the beginning of each year for $10$ years. Fund $1$ pays interest annually into Fund $2$. Fund $1$ earns $7$% annually while Fund $2$ earns $6$% annually. After $10$ years, Thomas has a total of $50000$. Calculate X. Interest earned is given by $0.07 X$ Amount at the end of $10$ years $50000=X+0.07X(\frac{1.06^{10}-1}{0.06})$ This gives $X=26 005.69$, when answer is $3416.80$

The equation is

$$10X+0.07\cdot X\cdot \sum_{k=1}^{10} \frac{1-1.06^k}{1-1.06}=50,000$$

You have already found out that $X$ is invested ten times.

The last period one interest payment is made due the last investment, but not compounded ($k=0$, the fraction is 1).

The period before another investment is made. The interest for the last period are $0.07X$ (not compounded). The interest of the interest of the period before are $0.07X\cdot 1.06$ ($k=1$). You can go on backward like this.

The equation can be simplified to

$$10X+\frac{0.07}{0.06}\cdot X\cdot (1.06\cdot \frac{1.06^{10}-1}{0.06}-10)=50,000$$