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Explanation:

S = i=01(2i+1)(2i+2)(2i+3)

12 i=0(2i+3)-(2i+1)(2i+1)(2i+2)(2i+3)

12 i=01(2i+1)(2i+2)-1(2i+2)(2i+3)

12i=012i+1-22i+2+12i+3

12i=012i+1+12i+3 - i=012i+2            ...(i)

Now, 12i=012i+1+12i+3 = 1211+13+13+15+15+17+17+...

121+213+15+17+...

12211+13+15+17+...-1

11+13+15+17+... - 12

= -12 + i=0 12i+1                                         ...(ii)

From (i) and (ii), we get,

S = -12 + i=012i+1-12i+2

= -1211-12+13-14+15-16+...

= -12 + i=0-1(i-1) 1i                                ...(iii)

Now, we know that,

log(1 + x) = x - 12x2 + 13x3 - 14x4  + ...

i=0(-1)n- 1  1nxn 

Putting x = 1 in the above equation, we get,

log2 = i=0-1(i-1) 1i                            ...(iv)

From (iii) and (iv), we get,

S = log2 - 12

Hence, option (d).

Alternatively,

You can also solve this question using options.

11.2.3 + 13.4.5 + 15.6.7 + ... = 16 + 160 + 1210 + ...

= 0.167 + 0.0167 + 0.0047 + …

< 0.2 but always positive

So, value of S will always be less than 0.2.

Now, we will consider all the given options one by one.

Consider option 1.

e2 – 1 = (2.718)– 1 > 0.2

So, option 1 can be eliminated.

Consider option 2.

loge2 – 1 = 0.693 – 1 < 0

So, option 2 can be eliminated.

Consider option 3.

2log102 – 1 = 2(0.3010) – 1 < 0

So, option 3 can be eliminated.

So, the correct answer will be option 4.

Hence, option (d).

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