The sum of the first 20 terms of the series $5 + 11 + 19)n (+ 29 + 41 + ....... is

n (c): Let S=5+11+19+29+ * s ... .+T_n n => S=5+11+19+ * s ... ... +T_n-1+T_n n Subtracting above two equations, we get n 0=5+\6+8+10+12+ * s ... .(n-1) terms \- T_n n => T_n=5+2(3+4+5+ * s ... upto (-1) terms ) n =5+2(1+2+3+ * s . . upto (n+1) terms -1-2) n => T_n=5+2 * ((n+1)(n+2) / 2)-6=n^2+3 n+2-1 n => T_n=n^2+3 n+1 Therefore S_n=Sigma n^2+3 Sigma n+Sigma 1 n => (n(n+1)(2 n+1) / 6)+(3(n+1) n / 2)+n n => S_20=(20 x 21 x 41 / 6)+(3 x 20 x 21 / 2)+20=3520n\

The sum of the first 20 terms of the series 5 + 11 + 19... | Sequence & Series (Mathematics)

\begin{aligned}\n& \text { (c): Let } S=5+11+19+29+\cdots \ldots .+T_n \\\n& \Rightarrow S=5+11+19+\cdots \ldots \ldots+T_{n-1}+T_n \\\n& \text { Subtracting above two equations, we get } \\\n& 0=5+\{6+8+10+12+\cdots \ldots .(n-1) \text { terms }\}-\mathrm{T}_n \\\n& \Rightarrow T_n=5+2(3+4+5+\cdots \ldots \text { upto }(-1) \text { terms }) \\\n& =5+2(1+2+3+\cdots . . \text { upto }(n+1) \text { terms }-1-2) \\\n& \Rightarrow T_n=5+2 \cdot \frac{(n+1)(n+2)}{2}-6=n^2+3 n+2-1 \\\n& \Rightarrow T_n=n^2+3 n+1 \therefore S_n=\Sigma n^2+3 \Sigma n+\Sigma 1 \\\n& \Rightarrow \frac{n(n+1)(2 n+1)}{6}+\frac{3(n+1) n}{2}+n \\\n& \Rightarrow S_{20}=\frac{20 \times 21 \times 41}{6}+\frac{3 \times 20 \times 21}{2}+20=3520\n\end{aligned}

Explanation

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