$5^{th}$ degree polynomial expression 
$p(x)$ is a $5$ degree polynomial such that
$p(1)=1,p(2)=1,p(3)=2,p(4)=3,p(5)=5,p(6)=8,$ then $p(7)$

$\bf{My\; Try::}$ Here We can not write the given polynomial as $p(x)=x$
and $p(x)=ax^5+bx^4+cx^3+dx^2+ex+f$ for a very complex system of equation,
plz hel me how can i solve that question, Thanks
 A: Like this problem, using difference of differences method,

A: let $x_i=i\,$, $\,i=1,2,\cdots,6$ and apply Lagrange's interpolation method
$${{L}_{i}}(x)=\frac{\prod\limits_{j\ne i,j=1}^{6}{(x-{{x}_{j}})}}{\prod\limits_{j\ne i,j=1}^{6}{({{x}_{i}}-{{x}_{j}})}}\,\,\,\,,\,\,\,i=1,2,\ldots ,7$$
$$P(x)=\sum\limits_{i=1}^{6}{{{L}_{i}}}(x)P({{x}_{i}})$$
A: Let's do it in the most elementary way. Let $$Q(x)=P(x+1)-P(x)-x+2 \tag{1}$$Observe that $Q$ is of degree $4$ and $Q(3)=Q(4)=Q(5)=0$. Therefore we can write$$Q(x)=a(x-3)(x-4)(x-5)(x-b) \tag{2}$$You have also from $(1)$ that $Q(1)=Q(2)=1$, which after substitution in $(2)$ you get $a=-1/8$ and $b=2/3$. So $$Q(6)=-\frac{1}{8}(6-3)(6-4)(6-5)\left(6-\frac{2}{3} \right)=-4$$And finally$$P(7)=Q(6)+P(6)+6-2=-4+8+6-2=8$$
A: hint : write the polynomial in this form $$f(x)= a(x-1)(x-2)(x-3)(x-4)(x-5)+b(x-1)(x-2)(x-3)(x-4)(x-6) +c(x-1)(x-2)(x-3)(x-5)(x-6)+d(x-1)(x-2)(x-4)(x-5)(x-6)+e(x-1)(x-3)(x-4)(x-5)(x-6)+f(x-2)(x-3)(x-4)(x-5)(x-6)$$ now finding constants are easy 
A: HINT:
Let $$\dfrac{p(x)}{(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)}=\sum_{i=1}^6\dfrac{A_i}{x-i}$$
Multiply both sides by $(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)$ and put $x=1,2,3,4,5,6$ one by one in the resultant identity.
A: Assume $p(x)$ of the form
$$p(x)=a\prod_{r=1}^5(x-r)+b\prod_{r=1}^4(x-r)+c\prod_{r=1}^3(x-r)+d\prod_{r=1}^2(x-r)+e(x-1)+f$$ Now put the values of $x$ i.e. $x=1,2,3,4,5,6$ , then values of $a,b,c,d,e,f$ will be $[\frac{-1}{40},\frac{1}{12},\frac{-1}{6},\frac{1}{2},0,1]$ respectively. you can get these values very easily and with alomost no calculation. Start with $x=1$ and get the value  of $f$ and then put more values to get $b,c,d,e,f$
So  $p(7)=8$
Hope this will help as this method does not solves the complicated equations.
A: Substitute $x$ for the appropriate values in your expression $$p(x)=ax^5+bx^4+cx^3+dx^2+ex+f$$and you will have a system of linear equations in $6$ variables ($a,b,c,d,e,f)$ that can be dealt with your favorite method, including linear algebra tech.
For instance, $p(2)=1$ yields $32a+16b+8c+4d+2e+f=1$.
After solving said system, you will have an explicit formula for $p(x)$, and then all you have to do is plug $x=7$ into it to obtain $p(7)$.
