Question

If \( S_n = 4 + 11 + 21 + 34 + 50 + \dots \) to \( n \) terms, then \( \frac{1}{60} (S_{29} - S_9) \) is equal to:

Answer 1

We are given the sequence \( S_n = 4 + 11 + 21 + 34 + 50 + \dots \) up to \( n \) terms, and we are asked to find \( \frac{1}{60} (S_{29} - S_9) \). 

Identify the general term of the sequence:-
The sequence appears to be quadratic, so let's find the general term \( a_n \) for the sequence. The first few terms are: \[ 4, 11, 21, 34, 50, \dots \] 

The correct answer is 223
\(S=4+11+21+34+ ........ +T_n\)
\(S= \quad4+11+21+.....+Tn-1+Tn\\\overline{\quad Tn=4+7+10+13+.....+(Tn+Tn-1)}\)
\(T_n=4+\frac{(n-1)}{2} [ 14+(n-2)3]\)
\(\quad\ =4+\frac{(n-1)}{2} [8+3n]\)
\(T_n=4+\frac{1}{2} (3n^2+5n-8)\)
\(∑T_n = S_n=\frac{3}{2}∑n^2 + \frac{5}{2}∑n\)
\(\quad\quad\ \ =\frac{3}{2} \frac{n(n+1)(2n+1)}{6} + \frac{5}{2} \frac{n(n+1)}{2}\)
\(\quad\quad\ \ \ \frac{S_{29} - S_{9}}{60} = 223\)