(3) Find the digit at unit place in the number .

16 ¹⁹⁸⁶ + 11¹⁹⁸⁶ - 6¹⁹⁸⁶. ?

And Let E = 16 ¹⁹⁸⁶ + 11¹⁹⁸⁶ - 6¹⁹⁸⁶

So, E = (10 + 6)ⁿ + (1 + 10)ⁿ - 6 ⁿ

where N = 1986

= 10^N + ^N C₁ 10^N-1 6 + ..........+ ^NC_N-1 10. 6^n-1 + 6^N

+ 1 + ^N C₁ 10. + ^N C₂ 10² + .........+ ^N C_N10 ^N - 6^N

= 1 + All inteqers being multiple of 10

So, The digit at unit place is 1.

(4) Find the value of x for which the sixth term of is equal to 21 and binomial coefficient of second, third and fourth terms are first, third and fifth terms of an aritnmetic progression .

Ans The sixth term of the given binomial exponsion is

^mC₅ ..................(1)

The other given lcondition is, m_C1 + m_C3 = 2 m_C2

=> m² - 9m + +14 = 0 => (m - 7) (m - 2) = 0

or m = 7 Since m = 2 is not possible .

Put m = 7 in equation (1) to get x = 0, 2

Dumb Question Why m cannot be 2 ?

If m = 2 then there cannot be 6^th term in the expansion and hence it is ruled out .

If n is any positive integer show that . 2³ⁿ⁺³ - 7n - 8 is divisible by 49 ?

The given expression

= 2³ⁿ⁺³ - 7n - 8

= 2³⁽ⁿ⁺¹⁾ - 7n - 8

= 8 ⁿ⁺¹ - 7n - 8

=(1 + 2)ⁿ⁺¹ - 7n - 8

= 8 (1 + 7) ⁿ- 7n - 8

= 8 (1 + ⁿC₁ 7 + ⁿ C₂. 7²........ⁿC_n 7ⁿ ) - 7_n - 8

= 8 + 56_n + 8 (ⁿC₂ 7² ........+ ⁿ C_n 7 ⁿ ) - 7_n- 8

= 49 _n 8 (ⁿ C₂ 7² +..............+ ⁿ C_{n 7ⁿ} )

= 49 (n + 8 ( ⁿ C₂ + .......+ ⁿ C_n 7 ^n-2 ) )

So, 2³ⁿ⁺³ - 7_n - 8 is divisible by 49 .

(6) Find the coefficient of x ⁵⁰ in the polynoamial

(1 + x) + 2(1 + x)² + 3(1 + x)³+.........+1000 (1 + x) ¹⁰⁰⁰ ?

Ans Let P(x) = (1 + x) + 2(1 + x)² + 3(1 + x)³+.........+1000 (1 + x) ¹⁰⁰⁰

Now (1 + x) p(x) - P(x)

= (1 + x)² + 2(1 + x)³ + 3(1 + x)⁴ +.........+999 (1 + x) ¹⁰⁰⁰+ 1000(1 + x) ¹⁰⁰⁰

- (1 + x) - 2 (1 + x)² - 3 (1 + x)³ ....................- 1000 (1 + x)¹⁰⁰⁰

= 1000 (1 + x)¹⁰⁰¹ - (1 + x) - (1 + x)² - (1 + x)³-.................-(1 + x)¹⁰⁰⁰

1000 (1 + x) ¹⁰⁰¹ - [ (1 + x)- (1 + x)² +.........+ (1 + x)¹⁰⁰⁰]

= 1000 (1 + x)¹⁰⁰¹-

= 1000 (1 + x) ¹⁰⁰¹ -

= P(x) =

So, coefficient of x⁵⁰ in P(x)

= 1000 X ¹⁰⁰⁰ C₅₁ = ¹⁰⁰⁰C₅₂

=

=

(7) If (1 + x )ⁿ = a_o + a₁x + a₂x²+ ..........+a_n xⁿ then find

value of ?

Ans Clearly a_r = ⁿ C_r

So,

=> 1 +

So,

Sum the following series

C_o + 5c₁ + ac₂ + ac₂+.........................+(4n + 1)c_n

Ans Let S = C_o + 5c₁ + ac₂ + ac₂+.........................+(4n + 1)c_n............(1)

Sine C_r = C_n -r

So, S C_n + 5C_n-1 + ^a C_n +.............+ (4n = 1)C_o

or S = (4n + 1) C_o + (4n - 3) C₁+..................+5C_n-1+c_n......(2) Now Adding (i) + (2) we get

25 = (4n + 2) C_o + C₁ +............C_n )

=> S = (2n + 1) 2ⁿ

So, C_o + 5C₁ + 9C₂ + ........+ (4n + 1) C_n = (2n + 1) 2ⁿ