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PSLE Foundation Mathematics

<ol class="abc-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">Find the term independent of $x$ in the expansion of $\left(2x^2 + \frac{3}{x}\right)^9$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><ol class="i-list"><li data-list-item-id="subitem-1781511615480-i4ty918y5" data-list-item-path="item-2/subitem-1781511615480-i4ty918y5"><div class="list-item-paragraph">Find the first three terms of the expansion of $(2 - 3x)^5$ in ascending powers of $x$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="subitem-1781511646569-ijhb25enu" data-list-item-path="item-2/subitem-1781511646569-ijhb25enu"><div class="list-item-paragraph">Hence, find the value of $n$ such that the coefficient of $x$ in the expansion of $ (2 - 3x)^5(1 - 3x)^n$ is $-720$<span class="inline-marks">[3]</span></div></li></ol></li></ol>

Binomial Expansion for Positive Integer Powers

Coefficients in Products of Binomial Expansions

General Terms and Coefficients in Binomial Expansions

Finding Parameters in Binomial Expansions

2024 GESS PRELIMS P1 Q5

Binomial Theorem

<ol class="abc-list"><li data-list-item-id="item-5-a" data-list-item-path="item-5-a"><div class="list-item-content"><span class="list-item-text">By considering the general term in the binomial expansion of $\left(x^4-\frac{1}{kx^2}\right)^6$, where&nbsp;$k$ is a positive constant, explain why there are only even powers of $x$ in this expansion.</span><span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-5-b" data-list-item-path="item-5-b"><div class="list-item-content"><span class="list-item-text">Given that the term independent of $x$ in the binomial expansion of $\left(x^4-\frac{1}{kx^2}\right)^6$ is $\frac{5}{27}$, show that&nbsp;$k$ is $3$.</span><span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1772870286453-hmj11qotg" data-list-item-path="item-1772870286453-hmj11qotg"><div class="list-item-paragraph">Hence find the term independent of $x$ in $ \left(2-3x^6 \right) \left(x^4-\frac{1}{kx^2} \right)^6$.<span class="inline-marks">[4]</span></div></li></ol>

By considering the general term in the binomial expansion of $\left(x^4-\frac{1}{kx^2}\right)^6$, where $k$ is a positive constant, explain why there are only even powers of $x$ in this expansion.

Given that the term independent of $x$ in the binomial expansion of $\left(x^4-\frac{1}{kx^2}\right)^6$ is $\frac{5}{27}$, show that $k$ is 3.

Hence find the term independent of $x$ in $(2-3x^6)\left(x^4-\frac{1}{kx^2}\right)^6$.

2025 SCGS PRELIMS P2 Q6

<p>In the expansion of $\left(x^2 + \frac{1}{3x}\right)^n$ in descending powers of $x$, the coefficients of the second and the third term are in the ratio of $6 : 11$.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Show that the value of $n$ is $12$.</span><span class="inline-marks">[6]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the middle term.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the coefficient of $x^6$ in the expansion of $(2 - x^3)\left(x^2 + \frac{1}{3x}\right)^{12}$.</span><span class="inline-marks">[3]</span></div></li></ol>

In the expansion of $\left(x^2 + \frac{1}{3x}\right)^n$ in descending powers of $x$, the coefficients of the second and the third term are in the ratio of $6 : 11$.

Find the coefficient of $x^6$ in the expansion of $(2 - x^3)\left(x^2 + \frac{1}{3x}\right)^{12}$.

2023 Ahmad Ibrahim Sec AM S3 EOY Q11 

<ol class="i-list"><li><div class="list-item-paragraph">Find the first three terms in the expansion, in ascending powers of $x$, of ${{\left( 2+x \right)}^{8}}$.</div></li><li><div class="list-item-paragraph">Hence, find the term in ${{x}^{2}}$ in the expansion of ${{\left( 2+x-{{x}^{2}} \right)}^{8}}$.</div></li></ol>

<p>In the expansion ${{\left( 2{{x}^{2}}+\frac{3}{x} \right)}^{n}}$, in descending powers of $x$, the ratio of the coefficients of the third and first term is $81:1$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Find the value of $n$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Write down the first three terms of the expansion.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Find the term that is independent of $x$.<span class="inline-marks">[2]</span></div></li></ol>

In the expansion ${{\left( 2{{x}^{2}}+\frac{3}{x} \right)}^{n}}$, in descending powers of $x$, the ratio of the coefficients of the third and first term is $81:1$.

2016 HIHS AM P2 Q3

<p>In the expansion of ${{\left( ax+\frac{1}{x} \right)}^{n}}$ in descending powers of $x$, where&nbsp;$a$ and $n$ are positive integers, the fourth term of the expansion is the constant term.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Find the value of $n$ and hence, express the constant term in terms of $a$.<span class="inline-marks">[4]</span></div></li><li><div class="list-item-paragraph">Using your value of $n$ in <strong>(i)</strong>, determine if a term in $\frac{1}{{{x}^{3}}}$ in the expansion $\left( 1-x \right){{\left( ax+\frac{1}{x} \right)}^{n}}$ exists.<span class="inline-marks">[2]</span></div></li></ol>

In the expansion of ${{\left( ax+\frac{1}{x} \right)}^{n}}$ in descending powers of $x$, where $a$ and $n$ are positive integers, the fourth term of the expansion is the constant term.

2016 VS AM P1 Q4

<ol class="i-list"><li><div class="list-item-paragraph">Find the constant term in the binomial expansion of ${{\left( x-\frac{1}{2{{x}^{3}}} \right)}^{8}}.$</div></li><li><div class="list-item-paragraph">Given that the coefficient of $\frac{1}{{{x}^{4}}}$ in the expansion of ${{\left( 1+\frac{4}{{{x}^{2}}} \right)}^{n}}+{{\left( x-\frac{1}{2{{x}^{3}}} \right)}^{8}}$is $1049.$ Find $n$, where $n&gt;0$.</div></li></ol>

<p>Show that the binomial expansion ${{\left( x-\frac{1}{2{{x}^{3}}} \right)}^{15}}$does not have an independent term.</p>

Show that the binomial expansion ${{\left( x-\frac{1}{2{{x}^{3}}} \right)}^{15}}$does not have an independent term.

<p>Find the term independent of $x$ in the expansion of $x{{\left( 2x-\frac{1}{2{{x}^{2}}} \right)}^{8}}.$</p>

Find the term independent of $x$ in the expansion of $x{{\left( 2x-\frac{1}{2{{x}^{2}}} \right)}^{8}}.$

<p>In the expansion of ${{\left( {{x}^{4}}-\frac{2}{{{x}^{2}}} \right)}^{9}}$, find the coefficient of $\frac{1}{{{x}^{12}}}$.</p>

In the expansion of ${{\left( {{x}^{4}}-\frac{2}{{{x}^{2}}} \right)}^{9}}$, find the coefficient of $\frac{1}{{{x}^{12}}}$.

<p>Find the term which is independent of $x$ in the expansion of ${{\left( {{x}^{2}}-\frac{1}{2{{x}^{6}}} \right)}^{12}}$.</p>

Find the term which is independent of $x$ in the expansion of ${{\left( {{x}^{2}}-\frac{1}{2{{x}^{6}}} \right)}^{12}}$.

<p>Find the term independent of $x$ in the expansion of ${{\left( \frac{1}{x}-\frac{{{x}^{2}}}{2} \right)}^{12}}$.</p>

Find the term independent of $x$ in the expansion of ${{\left( \frac{1}{x}-\frac{{{x}^{2}}}{2} \right)}^{12}}$.

<ol class="i-list"><li><div class="list-item-paragraph">Write down the general term in the binomial expansion of ${{\left( \frac{{{x}^{2}}}{4}+\frac{2}{x} \right)}^{18}}.$</div></li><li><div class="list-item-paragraph">Write down the power of $x$ in this general term.</div></li><li><div class="list-item-paragraph">Hence, or otherwise, determine the term independent of $x$ in the binomial expansion of ${{\left( \frac{{{x}^{2}}}{4}+\frac{2}{x} \right)}^{18}}.$</div></li></ol>

<p>Find the term independent of $x$ in the expansion of ${{\left( {{x}^{3}}-\frac{1}{2x} \right)}^{12}}$.</p>

Find the term independent of $x$ in the expansion of ${{\left( {{x}^{3}}-\frac{1}{2x} \right)}^{12}}$.

<p>In the expansion of ${\left( x^2 + \frac{1}{3x} \right)^n}$ in descending powers of ${x}$, the coefficients of the second and the third term are in the ratio of ${6:11}$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Show that the value of ${n}$ is ${12}$.<span class="inline-marks">[6]</span></div></li><li><div class="list-item-paragraph">Find the middle term.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Find the coefficient of ${x^6}$ in the expansion of ${\left(2 - x^3\right)\left(x^2 + \frac{1}{3x}\right)^{12}}$.<span class="inline-marks">[3]</span></div></li></ol>

In the expansion of ${\left( x^2 + \frac{1}{3x} \right)^n}$ in descending powers of ${x}$, the coefficients of the second and the third term are in the ratio of ${6:11}$.

2023 AISS AM S3 EOY Q11

<p>Find the term independent of $x$ in the expansion ${{\left( {{x}^{2}}-\frac{1}{4{{x}^{2}}} \right)}^{8}}$.</p>

Find the term independent of $x$ in the expansion ${{\left( {{x}^{2}}-\frac{1}{4{{x}^{2}}} \right)}^{8}}$.

<p>Find the term independent of $x$ in the expansion ${{\left( 2x-\frac{1}{3x} \right)}^{10}}$.</p>

Find the term independent of $x$ in the expansion ${{\left( 2x-\frac{1}{3x} \right)}^{10}}$.

<ol class="abc-list"><li><div class="list-item-paragraph">By considering the general term in the binomial expansion of ${{\left( 2{{x}^{2}}-\frac{1}{x} \right)}^{7}}$, explain why there are only three terms of negative powers of $x$ in this expansion.</div></li><li><div class="list-item-paragraph">Find the term independent of $x$ in the expansion of ${{\left( 3x-{{x}^{2}} \right)}^{2}}{{\left( 2{{x}^{2}}-\frac{1}{x} \right)}^{7}}$.</div></li></ol>

By considering the general term in the binomial expansion of ${{\left( 2{{x}^{2}}-\frac{1}{x} \right)}^{7}}$, explain why there are only three terms of negative powers of $x$ in this expansion.

Find the term independent of $x$ in the expansion of ${{\left( 3x-{{x}^{2}} \right)}^{2}}{{\left( 2{{x}^{2}}-\frac{1}{x} \right)}^{7}}$.

<ol class="abc-list"><li><div class="list-item-paragraph">By considering the general term in the binomial expansion of ${{\left( \frac{3}{{{x}^{5}}}-\frac{x}{2} \right)}^{6}}$, explain why there are only two of non-negative powers of $x$ in this expansion.</div></li><li><div class="list-item-paragraph">Find the term independent of $x$ in the expansion of $\left( 6-\frac{5}{{{x}^{2}}} \right){{\left( \frac{3}{{{x}^{5}}}-\frac{x}{2} \right)}^{6}}$.</div></li></ol>

By considering the general term in the binomial expansion of ${{\left( \frac{3}{{{x}^{5}}}-\frac{x}{2} \right)}^{6}}$, explain why there are only two of non-negative powers of $x$ in this expansion.

Find the term independent of $x$ in the expansion of $\left( 6-\frac{5}{{{x}^{2}}} \right){{\left( \frac{3}{{{x}^{5}}}-\frac{x}{2} \right)}^{6}}$.

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