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1

### AIEEE 2004

The sum of the first n terms of the series $${1^2} + {2.2^2} + {3^2} + {2.4^2} + {5^2} + {2.6^2} + ....\,is\,{{n{{(n + 1)}^2}} \over 2}$$ when n is even. When n is odd the sum is
A
$${\left[ {{{n(n + 1)} \over 2}} \right]^2}$$
B
$${{{n^2}(n + 1)} \over 2}$$
C
$${{n{{(n + 1)}^2}} \over 4}$$
D
$$\,{{3n(n + 1)} \over 2}$$

## Explanation

If $$n$$ is odd, the required sum is

$${1^2} + {2.2^2} + {3^2} + {2.4^2} + ......$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 2.{\left( {n - 1} \right)^2} + {n^2}$$

$$= {{\left( {n - 1} \right){{\left( {n - 1 + 1} \right)}^2}} \over 2} + {n^2}$$

[ As $$\left( {n - 1} \right)$$ is even

$$\therefore$$ using given formula for the sum of $$\left( {n - 1} \right)$$ terms.]

$$= \left( {{{n - 1} \over 2} + 1} \right){n^2} = {{{n^2}\left( {n + 1} \right)} \over 2}$$
2

### AIEEE 2004

Let $${{T_r}}$$ be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n, $$m \ne n,\,\,{T_m} = {1 \over n}\,\,and\,{T_n} = {1 \over m},\,$$ then a - d equals
A
$${1 \over m} + {1 \over n}$$
B
1
C
$${1 \over {m\,n}}$$
D
0

## Explanation

$${T_m} = a + \left( {m - 1} \right)d = {1 \over n}...........\left( 1 \right)$$

$${T_n} = a + \left( {n - 1} \right)d = {1 \over m}..........\left( 2 \right)$$

$$\left( 1 \right) - \left( 2 \right) \Rightarrow \left( {m - n} \right)d$$

$$= {1 \over n} - {1 \over m} \Rightarrow d = {1 \over {mn}}$$

From $$\left( 1 \right)$$ $$a = {1 \over {mn}} \Rightarrow a - d = 0$$
3

### AIEEE 2003

The sum of the serier $${1 \over {1.2}} - {1 \over {2.3}} + {1 \over {3.4}}..............$$ up to $$\infty$$ is equal to
A
$$\log {\,_e}\left( {{4 \over e}} \right)\,\,$$
B
$$2\,\log {\,_e}2$$
C
$$\log {\,_e}2 - 1\,$$
D
$$\log {\,_e}2$$

## Explanation

$${1 \over {1.2}} - {1 \over {2.3}} + {1 \over {3.4}}..........\infty$$

$$\left| {{T_n}} \right| = {1 \over {n\left( {n + 1} \right)}} = \left( {{1 \over n} - {1 \over {n + 1}}} \right)$$

$$S = {T_1} - {T_2} + {T_3} - {T_4} + {T_5}.........\infty$$

$$= \left( {{1 \over 1} - {1 \over 2}} \right) - \left( {{1 \over 2} - {1 \over 3}} \right) + \left( {{1 \over 3} - {1 \over 4}} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( {{1 \over 4} - {1 \over 5}} \right).......$$

$$= 1 - 2\left[ {{1 \over 2} - {1 \over 3} + {1 \over 4} - {1 \over 5}.........\infty } \right]$$

$$= 1 - 2\left[ { - \log \left( {1 + 1} \right) + 1} \right]$$

$$= 2\log 2 - 1 = \log \left( {{4 \over e}} \right)$$
4

### AIEEE 2002

$${1^3} - \,\,{2^3} + {3^3} - {4^3} + ... + {9^3} =$$
A
425
B
- 425
C
475
D
- 475

## Explanation

$${1^3} - {2^3} + {3^3} - {4^3} + ...... + {9^3}$$

$$= {1^3} + {2^3} + {3^3} + ...... + {9^3}$$

$$\,\,\,\,\,\,\,\,\,\, - 2\left( {{2^3} + {4^3} + {6^3} + {8^3}} \right)$$

$$= {\left[ {{{9 \times 10} \over 2}} \right]^2} - {2.2^3}\left[ {{1^3} + {2^3} + {3^3} + {4^3}} \right]$$

$$= {\left( {45} \right)^2} - 16.{\left[ {{{4 \times 5} \over 2}} \right]^2}$$

$$= 2025 - 1600 = 425$$

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