If - - are the roots of the equation x 2-4 x - p -0 where p - r -0 n n C r 1- rt -1- nt r -1 r- then - is

P=∑^n_r=0 ^nC_r1+rt/(1+nt)^r· ( 1)^r =∑^n_r=0 ^nC_r1/(1+nt)^r· ( 1)^r+ nbsp;nt∑^n_r=1 ^n 1C_r 11/(1+nt)^r· ( 1)^r =(1 1/1+nt)^n nt/1+nt∑^n_r=0 ^n 1C_r 1( 1 ...

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

If α, β are t...

Question

If $α, β$ are the roots of the equation $x^{2} + 4 x + p = 0$ where $p = n \sum r = 0^{n} C_{r} \frac{1 + r t}{(1 + n t)^{r}} \cdot (- 1)^{r}$ , then $| α - β |$ is

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Solution

$p = n \sum r = 0^{n} C_{r} \frac{1 + r t}{(1 + n t)^{r}} \cdot (- 1)^{r} = n \sum r = 0^{n} C_{r} \frac{1}{(1 + n t)^{r}} \cdot (- 1)^{r} + n t n \sum r = 1^{n - 1} C_{r - 1} \frac{1}{(1 + n t)^{r}} \cdot (- 1)^{r} = {(1 - \frac{1}{1 + n t})}^{n} - \frac{n t}{1 + n t} n \sum r = 0^{n - 1} C_{r - 1} {(\frac{- 1}{1 + n t})}^{r - 1} = {(\frac{n t}{1 + n t})}^{n} - (\frac{n t}{1 + n t}) {(1 - \frac{1}{1 + n t})}^{n - 1} = 0$

So, $x^{2} + 4 x = 0 \Rightarrow x = 0, - 4$
So, $| α - β | = 4$

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If α,β are the roots of the equation x2+4x+p=0 where p=n∑r=0 nCr1+rt(1+nt)r⋅(−1)r, then |α−β| is

p=n∑r=0 nCr1+rt(1+nt)r⋅(−1)r=n∑r=0 nCr1(1+nt)r⋅(−1)r+ntn∑r=1 n−1Cr−11(1+nt)r⋅(−1)r=(1−11+nt)n−nt1+ntn∑r=0 n−1Cr−1(−11+nt)r−1=(nt1+nt)n−(nt1+nt)(1−11+nt)n−1=0 So, x2+4x=0⇒x=0,−4 So, |α−β|=4

If $α, β$ are the roots of the equation $x^{2} + 4 x + p = 0$ where $p = n \sum r = 0^{n} C_{r} \frac{1 + r t}{(1 + n t)^{r}} \cdot (- 1)^{r}$ , then $| α - β |$ is