Question
The sum of n terms of the series $4+44+444+....$ is
The correct answer is
$(4/81)\ [10^{n+1}\ –\ 9n\ –\ 10]$
To find the sum of the series $4+44+444+....$ up to n terms, we first rewrite the terms:
- The k-th term, $T_k$, can be expressed as $4 \times (111....1)$ (k times).
- The number $111....1$ (k times) can be written as $\frac{10^k - 1}{9}$.
- Therefore, the k-th term is $T_k = 4 \times \frac{10^k - 1}{9}$.
Deriving the Sum Formula
The sum of the first n terms, $S_n$, is given by:
$S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \left( 4 \times \frac{10^k - 1}{9} \right)$Factor out the constant term:
$S_n = \frac{4}{9} \sum_{k=1}^{n} (10^k - 1)$Separate the summation:
$S_n = \frac{4}{9} \left( \sum_{k=1}^{n} 10^k - \sum_{k=1}^{n} 1 \right)$Evaluating the Summations
- The sum $\sum_{k=1}^{n} 1 = n$.
- The sum $\sum_{k=1}^{n} 10^k$ is a geometric series $10 + 10^2 + ... + 10^n$.
- First term $a = 10$.
- Common ratio $r = 10$.
- Number of terms = $n$.
- The sum is $a \frac{r^n - 1}{r-1} = 10 \frac{10^n - 1}{10-1} = 10 \frac{10^n - 1}{9} = \frac{10^{n+1} - 10}{9}$.
Final Sum Calculation
Substitute the evaluated sums back into the equation for $S_n$:
$S_n = \frac{4}{9} \left( \frac{10^{n+1} - 10}{9} - n \right)$Combine the terms inside the parenthesis over a common denominator:
$S_n = \frac{4}{9} \left( \frac{10^{n+1} - 10 - 9n}{9} \right)$Simplify the expression:
$S_n = \frac{4}{81} (10^{n+1} - 9n - 10)$This result matches the formula in option 3.
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Important Questions from Series
- In the sequence 6, 9, 14, $x$, 30, 41, a possible value of $x$ is
Let $a_0 = 0$ and define $a_n = \frac{1}{2}(1 + a_{n-1})$ for all positive integers $n \ge 1$.
The least value of $n$ for which $|1 - a_n| < \frac{1}{2^{10}}$ is __________.
(Answer in integer)
- Calculate the reciprocal of the coefficient of $z^3$ in the Taylor series expansion of the function $f(z) = \sin(z)$ around $z = 0$. (Provide the answer as an integer.)
- Let $a_1 = 1$ and $a_n = a_{n-1} + 4$, $n \ge 2$. Then,
$\lim_{n\to\infty} \left[\frac{1}{a_1a_2} + \frac{1}{a_2a_3} + \dots + \frac{1}{a_{n-1}a_n}\right]$
is equal to ________ - Let $S(x) = a_0 + \sum_{n=1}^\infty(a_n \cos (n x) + b_n \sin (n x))$ be the Fourier series of the$2 \pi$ periodic function defined by $f(x) = x^2 + 4 \sin (x) \cos(x)$, $-\pi \le x \le \pi$. Then
$|\sum_{n=0}^\infty a_n - \sum_{n=1}^\infty b_n|$
is equal to ________
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