If $a_{1}, a_{2}, a_{3}, . . . .$ are in A.P. then show that
$a_{1} C_{0} - a_{2} C_{1} + a_{3} C_{2} - . . . . . . + (- 1)^{n} a_{n + 1} C_{n} = 0$
Hence deduce that
$3 C_{0} - 8 C_{1} + 13 C_{2} - 18 C_{3} + . . . . = 0$

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Solution

Let $a_{1} = a a_{2} = a + d, . . . ., a_{n + 1} = a + n d .$
$L . H . S = a C_{0} - (a + d) C_{1} + (a + 2 d) C_{2} - . . . . . . + (- 1)^{n} (a + n d) C_{n}$
$= a {C_{0} - C_{1} + C_{2} - . . . .} - d {C_{1} - 2 C_{2} + 3 C_{3} - . . . .}$
$= a 0 - d . 0, = 0$ by Q. 1
Deduction : 3, 8 ,13 ,18,......form an A . P whose $a = 3, d = 5$
Putting $a = 3, d = 5$ we get the result $3.0 - 5.0 = 0$

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If a1,a2,a3,.... are in A.P. then show that a1C0−a2C1+a3C2−......+(−1)nan+1Cn=0Hence deduce that3C0−8C1+13C2−18C3+....=0

Let a1=aa2=a+d,....,an+1=a+nd.L.H.S=aC0−(a+d)C1+(a+2d)C2−......+(−1)n(a+nd)Cn=a{C0−C1+C2−....}−d{C1−2C2+3C3−....}=a0−d.0,=0 by Q. 1Deduction : 3, 8 ,13 ,18,......form an A . P whose a=3,d=5Putting a=3,d=5 we get the result 3.0−5.0=0

If $a_{1}, a_{2}, a_{3}, . . . .$ are in A.P. then show that
$a_{1} C_{0} - a_{2} C_{1} + a_{3} C_{2} - . . . . . . + (- 1)^{n} a_{n + 1} C_{n} = 0$
Hence deduce that
$3 C_{0} - 8 C_{1} + 13 C_{2} - 18 C_{3} + . . . . = 0$