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7.Binomial Theorem
hard
જો $a$ અને $d$ બે સંકર સંખ્યા હોય તો શ્રેણી $a{C_0} - (a + d){C_1} + (a + 2d){C_2} - ........$ ના $(n + 1)$ પદનો સરવાળો મેળવો.
A
$\frac{a}{{{2^n}}}$
B
$na$
C
$0$
D
એકપણ નહિ.
Solution
(c) We can write
$a{C_0} – (a + d)\,{C_1} + (a + 2d){C_2} – ….$upto $(n + 1)$terms
$ = a({C_0} – {C_1} + {C_2} – ….) + d( – {C_1} + 2{C_2} – 3{C_3} + ….)$ ….$(i)$
Again,${(1 – x)^n} = {C_0} – {C_1}x + {C_2}{x^2} – …. + {( – 1)^n}{C_n}{x^n}$ …$(ii)$
Differentiating with respect to $x$ ,$ – n{(1 – x)^{n – 1}} = – {C_1} + 2{C_2}x – …. + {( – 1)^n}{C_n}n{x^{n – 1}}$ ….$(iii)$
Putting $x =1$ in $(ii)$ and $(iii)$, we get ${C_0} – {C_1} + {C_2} – …. + {( – 1)^n}{C_n} = 0$ and $ – {C_1} + 2{C_2} – …. + {( – 1)^n}n.{C_n} = 0$
Thus the required sum to $(n+1)$ terms, by $(i) =a.0 + d.0 = 0.$
Standard 11
Mathematics
