SRCC GBO 2011 Question Paper

The series can be broken as: $$\ \frac{\ 1}{2}+\ \ \frac{\ 3}{4}+\ \frac{\ 7}{8}+.......\ \ \frac{\ 1023}{1024}$$ or $$\left(1-\ \frac{\ 1}{2}\right)+\left(\ 1-\ \frac{\ 1}{4}\right)+......$$

This becomes: 1+1+1+1..... n times - { $$\ \frac{\ 1}{2}+\ \frac{\ 1}{2^2}+.......\ \ \frac{\ 1}{2^n}$$ }

Here, n = 10

Hence, the sum becomes 10 - (Sum of GP upto 10 terms with a common ratio of 0.5)

This gives the answer as: 10 - $$\ \frac{\ 1}{2}\times\ \ \frac{\ \ 1-\frac{\ 1}{2^{10}}}{1-\ \frac{\ 1}{2}}$$ or approximately 9 (slightly greater than 9)

Since the square root of a number is between 6 and 7.

So, number itself will lie between $$6^2 = 36$$ and $$7^2 = 49$$

Cube root of 36 = 3.3 (Approx)

Cube root of 49= 3.66 (Approx)

Hence, the cube root of the number will be between 3 and 4. 

$$3^{210} = (3^2)^{105} = 9^{105} < 17^{105} => 3^{210} < 17^{105}$$

$$7^{140} = (7^4)^{35} = 2401^{35}$$

$$17^{105} = (17^3)^{35} = 4913^{35} => 7^{140} < 17^{105}$$

$$17^{105} = (17^5)^{21} = 1419857‬^{21}$$

$$31^{84} = (31^4)^{21} = 923521^{21} => 31^{84} < 17^{105}$$

Hence, largest of the number = $$17^{105}$$

Let 18 consecutive natural number = a, a + 1, a + 2, a + 3, ........., a + 17.

Sum = $$\frac{n}{2}[2a + (n - 1)d]$$

= $$\frac{18}{2}[2a + (18 - 1)1]$$

= 9 (2a + 17)

= 18a + 153

Put a = 1, Sum = 18 + 153 = 171

Put a = 2, Sum = 36 + 153 = 189

Put a = 3, Sum =54 + 153 = 207

Put a = 4, Sum =72 + 153 = 225 [Required answer]

$$\frac{1}{1 + 1^2 + 1^4} + \frac{2}{1 + 2^2 + 2^4} + \frac{3}{1 + 3^2 + 3^4}+ ....... + \frac{99}{1 + 99^2 + 99^4}$$

= $$\frac{1}{3} + \frac{2}{21} + \frac{3}{91} + ....... +$$

= $$\frac{1}{1 \cdot 3} + \frac{2}{3 \cdot 7} + \frac{3}{7 \cdot 13} + ....... +$$

$$\frac{1}{1 \cdot 3}$$ this can be written as $$\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right)$$

$$ \frac{2}{3 \cdot 7}$$ this can be written as $$\frac{1}{2}\left(\frac{1}{3}-\frac{1}{7}\right)$$

$$\frac{3}{7 \cdot 13}$$ this can be written as $$\frac{1}{2}\left(\frac{1}{7}-\frac{1}{13}\right)$$

The whole expression can be written as

=$$\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right)$$+ $$\frac{1}{2}\left(\frac{1}{3}-\frac{1}{7}\right)$$+$$\frac{1}{2}\left(\frac{1}{7}-\frac{1}{13}\right)$$ +....

The terms get cancelled out and we will be remained with the first from the first term and the second from the last term.

The second term in each of these terms form a pattern.

$$(n+1)^2-(n+1)+1$$

For the first term n=1, $$(n+1)^2-(n+1)+1$$ = 3

For the second term n=2, $$(n+1)^2-(n+1)+1$$ = 7

For the second term n=3, $$(n+1)^2-(n+1)+1$$ = 13

For the 99th term n=99, $$(n+1)^2-(n+1)+1$$ = 9901.

The final value of the expression is $$\frac{1}{2}-\frac{1}{9901}$$

This will be in between 0.49 and 0.5

Going by the traditional method since the numbers are easy to calculate, the value of $$71^2 - 37^2 - 51$$ is 3621.

The prime factorisation of 3621 is $$\ 1\times\ 3\times\ 17\times\ 71$$

Hence, there are three prime numbers that divide the expression.

$$x^2 - x - 1 = 0$$

$$x^2 = 1 + x$$ Or $$x^2 - x = 1$$

$$= x^3 - 2x + 1 = x.x^2 - 2x + 1$$

$$=x.(1 + x) - 2x + 1$$

$$=x + x^2 - 2x + 1$$

$$=x^2 - x + 1$$

$$=1 + 1$$

= 2

$$120 = 2 * 2 * 2 * 3 * 5 = 2^3 * 3^1 * 5^1$$

Sum of factors = $$(2^0 + 2^1 + 2^2 + 2^3) * (3^0 * 3^1) * (5^0 * 5^1) = (1 + 2 + 4 + 8) * (1 + 3) * (1 + 5) = 15 * 4 * 6 = 360$$

Ssum of the reciprocals factors = $$\frac{360}{120} = 3$$