Find the sum - - n -1101-2 n -1-1-5 n -1

∑_n=1^10{ ( 1/2 )^n 1 + ( 1/5 )^n+1} ∑_n=1^10 ( 1/2 )^n 1 + ∑_n=1^10 ( 1/5 )^n+1 =1 + 1/2 + 1/2^2 + ..... + 1/5^2 + 1/5^3 + 1/5^4 + ..... = ( 1 1/2^10 )/1 ...

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Byju's Answer

Standard XII

Mathematics

Multinomial Expansion

Find the sum ...

Question

Find the sum : $\sum_{n = 1}^{10} {{(\frac{1}{2})}^{n - 1} + {(\frac{1}{5})}^{n + 1}}$

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Solution

$\sum_{n = 1}^{10} {{(\frac{1}{2})}^{n - 1} + {(\frac{1}{5})}^{n + 1}} \sum_{n = 1}^{10} {(\frac{1}{2})}^{n - 1} + \sum_{n = 1}^{10} {(\frac{1}{5})}^{n + 1} = 1 + \frac{1}{2} + \frac{1}{2^{2}} + . . . . . + \frac{1}{5^{2}} + \frac{1}{5^{3}} + \frac{1}{5^{4}} + . . . . . = \frac{(1 - \frac{1}{2^{10}})}{1 - \frac{1}{2}} + \frac{\frac{1}{5} (1 - \frac{1}{5^{10}})}{1 - \frac{1}{5}} = \frac{2^{10} - 1}{2^{9}} + \frac{5^{10} - 1}{4 \times 5^{11}}$

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Find the sum : ∑10n=1{(12)n−1+(15)n+1}

∑10n=1{(12)n−1+(15)n+1}∑10n=1(12)n−1+∑10n=1(15)n+1=1+12+122+.....+152+153+154+.....=(1−1210)1−12+15(1−1510)1−15=210−129+510−14×511

Find the sum : $\sum_{n = 1}^{10} {{(\frac{1}{2})}^{n - 1} + {(\frac{1}{5})}^{n + 1}}$