Sequence And Series Question 431
Question: The 20th terms of the series 2 + 3 + 5 +9+16 +……. is
Options:
A) 950
B) 975
C) 990
D) 1010
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let $ \begin{matrix} S_1=2+3+5+9+16+……..+x_{n} \\ S_1=2+3+5+9+……..,{x_{n-1}}+x_{n} \\ \begin{aligned} & _____________________________________ \\ & O=2+[1+2+4+7+….to(n-1),terms]-x_{n} \\ \end{aligned} \\ \end{matrix} $
$ \therefore ,x_{n}=2+[1+2+4+7+……to,(n-1),terms] $ Again let $ ,\begin{matrix} S_2=1+2+4+7+………+{t_{n-1}} \\ S_2=1+2+4+7……..+{t_{n-2}}+{t_{n-1}} \\ \begin{aligned} & ______________________________ \\ & O=1+[1+2+3+…..+(n-2)terms]-{t_{n-1}} \\ \end{aligned} \\ \end{matrix} $ $ {t_{n-1}}=1+\frac{(n-2)(n-1)}{2}=\frac{n^{2}-3n+4}{2} $
$ \therefore ,S_2=\sum\limits_{n=1}^{n-1}{{t_{n-1}}=\frac{1}{2}\Sigma n^{2}-\frac{3}{2}\Sigma n+2\Sigma 1} $ $ =\frac{1}{2}\frac{(n-1)n(2n-1)}{6}-\frac{3}{2}\frac{n(n-1)}{2}+2(n-1) $ $ =(n-1),[ \frac{2n^{2}-n}{12}-\frac{3n}{4}+2 ] $ $ =\frac{n-1}{12}[ 2n^{2}-n-9n+24 ] $ $ =\frac{n-1}{6}[ n^{2}-5n+12 ]=\frac{n^{3}-6n^{2}+17n-12}{6} $
$ \therefore ,x_{n}=2+S_2=2+\frac{n^{3}-6n^{2}+17n-12}{6} $ $ =\frac{n^{3}-6n^{2}+17n}{6} $ So, $ x_{20}=990 $
BETA
