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Let for n = 1, 2,……, 50, Sn be the sum of the infinite geometric progression whose first term is n2 and whose common ratio is  Then the value of 1 2 6 + ∑ n = 1 5 0 ( S n + 2 n + 1 − n − 1 )  is equal to…………..

0 4 Views | Posted 2 months ago
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  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    2 months ago

    S n = n 2 1 − 1 ( n + 1 ) 2 = n ( n + 1 ) 2 ( n + 2 ) = ( n + 1 ) 2 − 2 ( n + 1 ) + 2 − 2 ( n + 2 )

    1 2 6 + ∑ n = 1 5 0 ( S n + 2 ( n + 1 ) − ( n + 1 ) ) = 1 2 6 + ∑ n = 1 5 0 ( n + 1 ) 2 − 3 ( n + 1 ) + 2 + 2 ( 1 ( n + 1 ) − 1 ( n + 2 ) )

    = 1 2 6 + 4 5 5 2 5 − 3 × 1 3 2 5 + 2 × 5 0 + 2 ( 1 2 − 1 5 2 )

    = 1 2 6 + 4 1 5 5 0 + 1 0 0 + 1 − 1 2 6 = 4 1 6 5 1

Similar Questions for you

A line passes through (9, 0), making angle 30° with positive direction of x-axis. It is rotated by angle of 15° with respect to (9, 0). Then one of the equation of new line is
A
alok kumar singh

Eqn : y – 0 = tan45° (x – 9) Þ y = (x – 9)

Option (B) is correct

If two circle x2 + y2 = 4 and x2 + y2 – 4lx + 9 = 0 intersect at two distinct points, then find the range of l.
A
alok kumar singh

|r1 – r2| < c1c2 < r1 + r2

-> | 2 − 4 λ 2 − 9 | < | 2 λ | < 2 + 4 λ 2 − 9

| 2 λ | − 2 < 4 λ 2 − 9    

4 λ 2 + 4 − 8 | λ | < 4 λ 2 − 9

  λ > 1 3 8 ,   λ < − 1 3 8           

4 λ 2 − 9 > 0

λ > 3 2 ,   λ < − 3 2

λ ∈ ( − ∞ ,   − 1 3 8 ) ∪ ( 1 3 8 ,   ∞ )           

Now,

| 2 − 4 λ 2 − 9 | < | 2 λ |            

4 + 4 λ 2 − 9 − 4 4 λ 2 − 9 < 4 λ 2  

4 4 λ 2 − 9 > − 5 ⇒ λ ∈ R  

λ ∈ ( − ∞ ,   − 1 3 8 ) ∪ ( 1 3 8 ,   ∞ )  

The line passes through the centre of circle x2 + y2 – 16x – 4y = 0, it interacts with the positive coordinate axis at A & B. Then find the minimum value of OA + OB, where O is origin.
A
alok kumar singh

(y – 2) = m (x – 8)

⇒   x-intercept

⇒     ( − 2 m + 8 )

⇒   y-intercept

⇒   (–8m + 2)

⇒   OA + OB = − 2 m 2  + 8 – 8m + 2

f ' ( m ) = 2 m 2 − 8 = 0  

-> m 2 = 1 4

-> m = − 1 2

-> f ( − 1 2 ) = 1 8

->Minimum = 18

The image of the point (3, 5) in the line x – y + 1 = 0, lies on:
V
Vishal Baghel

Kindly consider the following figure

A man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of the line on the coordinate axes is 1 4 .  Three stones A, B and C are placed at the point (1, 1), (2, 2) and (4, 4) respectively. Then which of these stones is / are on the path of the man?
V
Vishal Baghel

According to question,

1 2 ( 1 a + 1 b ) = 1 4

1 a + 1 b = 1 2 . . . . . . . . ( i )

Equation of required line is x a + y b = 1

Obviously B (2, 2) satisfying condition (i)

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