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Byju's Answer

Standard VII

Mathematics

Perfect Square Numbers

Find the valu...

Question

Find the value of $a + a^{2} - a^{3} + a^{4} - a^{5}$ if $a = - 2$ .

- 22

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- 18

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32

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58

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Solution

The correct option is D $58$
We need to find value of $a + a^{2} - a^{3} + a^{4} - a^{5}$
Put the value of $a = - 2$
$\Rightarrow (- 2) + (- 2)^{2} - (- 2)^{3} + (- 2)^{4} - (- 2)^{5}$
$\Rightarrow - 2 + (4) - (- 8) + (16) - (- 32)$
$\Rightarrow - 2 + 4 + 8 + 16 + 32$
$\Rightarrow - 2 + 60 = 58$

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Similar questions

Q. If

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

then

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

Q. Suppose

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\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}

where

0 \leq a < j

for

j = 2, 4, 5, 6, 7.

The sum

a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}

Suppose $a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ are integers such that $\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4!} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}$ where 0 $\leq$ a< j for j= 2,4,5,6,7.

The sum $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$ is

Q. If

a_{1}, a_{2}, a_{3}, a_{4}, a_{5}

are in A.P. with common difference

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then find the value of

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a_{3} = 2.

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a_{3}

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Find the value of a+a2−a3+a4−a5 if a=−2.

The correct option is D 58We need to find value of a+a2−a3+a4−a5Put the value of a=−2 ⇒(−2)+(−2)2−(−2)3+(−2)4−(−2)5⇒−2+(4)−(−8)+(16)−(−32)⇒−2+4+8+16+32⇒−2+60=58

Find the value of $a + a^{2} - a^{3} + a^{4} - a^{5}$ if $a = - 2$ .

The correct option is D $58$
We need to find value of $a + a^{2} - a^{3} + a^{4} - a^{5}$
Put the value of $a = - 2$
$\Rightarrow (- 2) + (- 2)^{2} - (- 2)^{3} + (- 2)^{4} - (- 2)^{5}$
$\Rightarrow - 2 + (4) - (- 8) + (16) - (- 32)$
$\Rightarrow - 2 + 4 + 8 + 16 + 32$
$\Rightarrow - 2 + 60 = 58$