Question

Show that 11n+2 + 122n+1 is divisible by 133.

Solution

Correct option is

133

11n+2 + 122n+1 = 112. 11n + 12(144)n

Now 144 and 121 should be expression in terms of 133; 144 as (133 + 11) or 121 as (133 – 12)

                      = 121. 11n + 12(11 + 133)n

                      = 11n. 133 + terms containing 133 as a factor

Hence the expression is divisiable by 133.

SIMILAR QUESTIONS

Q1

Find the 11th in the expression of 

Q2

Find the term independent of x in 

Q3

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Q4

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Q5

If the sum of the coefficients in the expression of (1 + 2x)n is 6561. The greatest term in the expression at x = 1/2 is

Q6

The number of terms in the expression of (a + b + c)n, where n Ïµ N, is

Q7

The number of terms which are free from fractional powers in the expansion of (a1/5 + b2/3)45a ≠ b is

Q8

If n is an odd natural, then  equals

Q9
Q10

The coefficient of xm in (1 + x)r + (1 + x)r+1 + (1 + x)r+2 +….+ (1 + x)nr≤ m ≤ n is