NIELIT 2022 April Scientist B | Section B | Question: 48

Question

The particular solution of the recurrence relation $a_{r+2} – 4a_{r+1} + 4a_{r} = 2^{r}$ is:

• A. $r.2^{r}$
• B. $r(r-1)2^{r-1}$
• C. $r(r-1)2^{r-2}$
• D. $r(r-1)2^{r-3}$

Answer 1

The correct answer is option (D).

The Particular solution of the recurrence relation must satisfy the recurrence relation .

The given recurrence relation is

$a\underset{r+2}{}-4a\underset{r+1}{}+4a\underset{r}{}=2^{r}$………………...(1)

try with option (4)

$a\underset{n}{}=r(r-1)2^{r-3}$

$a\underset{n+1}{}=r(r+1)2^{r-2}$

$a\underset{n+2}{}=(r+2)(r+1)2^{r-1}$

putting the value of $a\underset{r+2}{},a\underset{r+1}{},a\underset{r}{}$ in L.H.S of (1)

=>$(r+2)(r+1)2^{r-1}-4r(r+1)2^{r-2}+4r(r-1)2^{r-3}$

=>$(r^{2}+3r+2)2^{r-1}-4(r^{2}+1)2^{r-2}+4(r^{2}-r)2^{r-3}$

=>$2^{r-3}\left \{ 4r^{2}+12r+8-8r^{2}-8r+4r^{2}-4r \right \}$

=>$2^{r-3}*8$

=>$2^{r}$    ……..(RHS)

so correct answer is (D)