with annual compounding:
$\Sigma_t[S_1r(1+g)^{t-1}(1+i)^{T-t}]$
with other compounding frequencies:
$\Sigma_t[\Sigma_x[\frac{S_1r(1+g)^{t-1}}{m}(1+\frac{i}{m})^{(m-x)}](1+\frac{i}{m})^{[m(T-t)]}]$
but that's probably what you referred to as sum of finite series... nevertheless, it's quite a short expression, I think, do you really need an integral here? Sorry for asking naively..
Derivation of the above:
First simplified, with annual payments.
Timeline:
$t = 0$ Lisa starts working
$t = 1$ Lisa gets her first annual salary, and makes the first payment into her savings account.
$t= T = 40$ Lisa retires. So she makes the last payment in $t=40$ (still one payment into the account from the last salary at the retirement date)
She works 40 years, makes 40 payments, and retires 39 years after the first payment.
The annual payment $P(t)$ is her savings rate $r=10%$ times her salary $S_t$ which is 40 K at the end of the first year, i.e. $S_1=40,000$ and then grows with $g=$5% per year, i.e.:
$P_t = S_1r(1+g)^{t-1}$
Payments into her savings account grow at an annual interest rate of $i=7$%, so the value at the retirement date ($T$) of a payment made in $t$ is:
$FV_{S_t}=S_tr(1+i)^{(T-t)}=S_1r(1+g)^{t-1}(1+i)^{T-t}$
The total value of the account is the sum over $t$ till $t=40$:
$\Sigma_tFV_{P_t}=$
$\Sigma_t[S_tr(1+i)^{(T-t)}]=$
$\Sigma_t[S_1r(1+g)^{t-1}(1+i)^{T-t}]=$
$\Sigma_t4,000(1.05)^{t-1}(1.07)^{40-t}$
with other compounding frequency:
number of compounding (sub)periods per year:
index for current subperiod (here months): $x$ $(1,2,...,12)$
$\Sigma_t[\Sigma_x[\frac{S_1r(1+g)^{t-1}}{m}(1+\frac{i}{m})^{(m-x)}](1+\frac{i}{m})^{[m(T-t)]}]$
Best Answer
The future value of a contribution $c$ compounded for time $t$ is $c(1.085)^t$
If there are multiple contributions then we sum the future value of each
$c_1(1.085)^{t_1} + c_2(1.085)^{t_2} + \cdots$
June makes 12 contributions that compound for $47, 46, \cdots, 35$ years respectively.
$\sum_\limits{n=35}^{47} 1000 (1.085)^n$
Jas makes 30 contributions that comound from $30$ to $1$ year
$\sum_\limits{n=35}^{47} 1500 (1.085)^n$
Now we use some trickery for the sum of a series.
$\sum_\limits{n=a}^{b} r^n = \frac {r^{b+1}- r^a}{1-r}$
June's account is worth $1000\frac {(1.085)^{48} - (1.085)^{35}}{0.085}$ Jas' account is worth $1500\frac {(1.085)^{31} - (1.085)}{0.085}$
But these results are slightly different from what you show as the answers.
Reverse engineering it:
$1000\frac {(1.085)^{47} - (1.085)^{35}}{0.085} = 339,789\\ 1500\frac {(1.085)^{30} - (1.085)^{0}}{0.085} = 186,322$
These would suggest that the account is evaluated when June and Jas are 64, and that June did not contribute when she was 30.