Starting with the series:
\[S(x) = (1 + x) + 2(1 + x)^2 + 3(1 + x)^3 + \dots + 60(1 + x)^{60}\]
Multiplying both sides by \( (1 + x) \), we get:
\[(1 + x)S = (1 + x) + 2(1 + x)^2 + 3(1 + x)^3 + \dots + 60(1 + x)^{61}\]
Now, subtracting \( S \) from \( (1 + x)S \), we obtain:
\[-xS = \frac{(1 + x)(1 + x)^{60} - 1}{x} - 60(1 + x)^{61}\]
Now, put \( x = 60 \):
\[-60S = \frac{61((61)^{60} - 1)}{60} - 60 \cdot (61)^{61}\]
Solving this equation gives:
\[S = 3660\]
The shaded region in the Venn diagram represents
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the river bank is _________ cm. (Take \( g = 10 \, {m/s}^2 \)).