Find the 7 th term of the series 1- 1 1-3 1-2 2 - 1 1-3-5 1-2-3 2 -

The correct option is A 16 7^th term=(1+2+3....7)^2(1+3+5+7....13) T_7=[72(1+7)]^272(1+13)=28×814 T_7=16 ...

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Byju's Answer

Standard XII

Mathematics

Arithmetic Progression

Find the 7 ...

Question

Find the $7^{t h}$ term of the series $1 + \frac{1}{(1 + 3)} {(1 + 2)}^{2} + \frac{1}{(1 + 3 + 5)} {(1 + 2 + 3)}^{2} + . . . . . . . .$ ,

16

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Solution

The correct option is A $16$
$7^{t h} t e r m = \frac{(1 + 2 + 3....7)^{2}}{(1 + 3 + 5 + 7....13)} T_{7} = \frac{[\frac{7}{2} (1 + 7)]^{2}}{\frac{7}{2} (1 + 13)} = \frac{28 \times 8}{14} T_{7} = 16$

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Similar questions

Q. For the series

S = 1 + \frac{1}{(1 + 3)} (1 + 2)^{2} + \frac{1}{(1 + 3 + 5)} (1 + 2 + 3)^{2} + \frac{1}{(1 + 3 + 5 + 7)} (1 + 2 + 3 + 4)^{2} + \dots

Q. For the series,

S = 1 + \frac{1}{(1 + 3)} (1 + 2)^{2} + \frac{1}{(1 + 3 + 5)} (1 + 2 + 3)^{2} + \frac{1}{(1 + 3 + 5 + 7)} (1 + 2 + 3 + 4)^{2} + \dots \dots

For the series $S = 1 + \frac{1}{(1 + 3)} (1 + 2)^{2} + \frac{1}{(1 + 3 + 5)} (1 + 2 + 3)^{2} + \frac{1}{(1 + 3 + 5 + 7)} (1 + 2 + 3 + 4)^{2} + . . . . .$

Q. For the series,

S = 1 + \frac{1}{(1 + 3)} (1 + 2)^{2} + \frac{1}{(1 + 3 + 5)} (1 + 2 + 3)^{2} + \frac{1}{(1 + 3 + 5 + 7)} (1 + 2 + 3 + 4)^{2} + . . . .

Q. For the series,

S = 1 + \frac{1}{(1 + 3)} (1 + 2)^{2} + \frac{1}{(1 + 3 + 5)} (1 + 2 + 3)^{2} + \frac{1}{(1 + 3 + 5 + 7)} (1 + 2 + 3 + 4)^{2} + \dots \dots

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Find the 7th term of the series 1+1(1+3)(1+2)2+1(1+3+5)(1+2+3)2+........,

The correct option is A 167thterm=(1+2+3....7)2(1+3+5+7....13)T7=[72(1+7)]272(1+13)=28×814T7=16

Find the $7^{t h}$ term of the series $1 + \frac{1}{(1 + 3)} {(1 + 2)}^{2} + \frac{1}{(1 + 3 + 5)} {(1 + 2 + 3)}^{2} + . . . . . . . .$ ,

The correct option is A $16$
$7^{t h} t e r m = \frac{(1 + 2 + 3....7)^{2}}{(1 + 3 + 5 + 7....13)} T_{7} = \frac{[\frac{7}{2} (1 + 7)]^{2}}{\frac{7}{2} (1 + 13)} = \frac{28 \times 8}{14} T_{7} = 16$